Question 15 Marks
The sum of the reciprocals of Meena's ages (in years) 3 years ago and 5 years hence is $\frac{1}{3}.$ Find her present age.
AnswerLet the present age of Meena be x years.
Then,
3 years ago, Meena's age = (x - 3) years.
5 years hence, Meena's age = (x + 5) years
It is given that
$\frac{1}{\text{x}-3}+\frac{1}{\text{x}+5}=\frac{1}{3}$
$\Rightarrow\frac{\text{x}+5+\text{x}-3}{(\text{x}-3)(\text{x}+5)}=\frac{1}{3}$
$\Rightarrow\frac{\text{2x}+2}{\text{x}^2+\text{2x}-15}=\frac{1}{3}$
$\Rightarrow 6x + 6 = x^2 + 2x - 15$
$\Rightarrow x^2 - 4x - 21 = 0$
$\Rightarrow x^2 - 7x + 3x - 21 = 0$
$\Rightarrow x(x - 7) + 3(x - 7) = 0$
$\Rightarrow (x - 7)(x + 3) = 0$
$\Rightarrow x - 7 = 0 or x + 3 = 0$
$\Rightarrow x = 7 or x = -3$
Since age cannot be negative, $\text{x}\neq-3.$
⇒ x = 7
Hence, Meena's present age is 7 years.
View full question & answer→Question 25 Marks
A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the size of the square by one student, he found that he was short of 25 students. Find the number of students.
AnswerLet there be x rows and number of student in each row be x.
Then, total number of students $= (x^2 + 24)$
$\Rightarrow x^2 + 24 = (x + 1)^2 - 25$
$\Rightarrow x^2 + 24 + x^2 + 1 + 2x - 25$
$\Rightarrow 2x = 48$
$\Rightarrow x = 24$
Hence total number of student
$= [(24)^2 + 24] = 567 + 24 = 600$
Total number of students is 600.
View full question & answer→Question 35 Marks
Solve the following equations by using the method of completing the square:
$3x^2 - 2x - 1 = 0$
Answer$3x^2 - 2x - 1 = 0$
$\Rightarrow 9x^2 - 6x - 3 = 0$ (Multiplying both sides by 3)
$\Rightarrow 9x^2 - 6x = 3$
$\Rightarrow (3x)^2 - 2 \times 3x \times 1 + 1^2 = 3 + 1^2$ [Adding $1^2$ on both sides]
$\Rightarrow (3x - 1)^2 = 3 + 1 = 4 = (2)^2$
$\Rightarrow\text{3x}-1=\pm2$ (Taking square root on both sides)
⇒ 3x - 1 = 2 or 3x - 1 = -2
⇒ 3x = 3 or 3x = -1
⇒ x = 1 or $\text{x}=-\frac{1}{3}$
Hence, 1 and $-\frac{1}{3}$ are the roots of the given equation.
View full question & answer→Question 45 Marks
Solve the following equations by using the method of completing the square:
$2x^2 + 5x - 3 = 0$
Answer$2x^2 + 5x - 3 = 0$
$\Rightarrow 4x^2 + 10x - 6 = 0$ (Multiplying both sides by 2)
$\Rightarrow 4x^2 + 10x = 6$
$\Rightarrow(\text{2x})^2+2\times\text{2x}\times\frac{5}{2}+\Big(\frac{5}{2}\Big)^2\\=6+\Big(\frac{5}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{5}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}+\frac{5}{2}\Big)^2$
$=6+\frac{25}{4}$
$=\frac{24+25}{4}=\frac{49}{7}=\Big(\frac{7}{2}\Big)^2$
$\Rightarrow\text{2x}+\frac{5}{2}=\pm\frac{7}{2}$ (Taking square root on both sides)
$\Rightarrow\text{2x}+\frac{5}{2}=\frac{7}{2}$ or $\text{2x}+\frac{5}{2}=-\frac{7}{2}$
$\Rightarrow\text{2x}=\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1$ or $\text{2x}=-\frac{7}{2}-\frac{5}{2}=-\frac{12}{2}=-6$
$\Rightarrow\text{x}=\frac{1}{2}$ or x = -3
Hence $\frac{1}{2}$ and -3 are the roots of the given equation.
View full question & answer→Question 55 Marks
The product of Tanvy's age (in years) 5 years ago and her age 8 years later is 30. Find her present age.
AnswerLet the present age of Tanvy be x years. Then$,(x - 5)(x + 8) = 30$
$\Rightarrow x^2 + 8x - 5x - 40 = 30$
$\Rightarrow x^2 + 3x - 40 - 30 = 0$
$\Rightarrow x^2 + 3x - 70 = 0$
$\Rightarrow x^2 + 10x - 7x - 70 = 0$
$\Rightarrow x(x + 10) - 7(x + 10) = 0$
$\Rightarrow (x + 10)(x - 7) = 0$
$\Rightarrow x + 10 = 0 or x - 7 = 0$
$\Rightarrow x = -10 or x = 7$
⇒ x = 7 $(\because$ age cannot be negative$)$
Hence, the present age of Tanvy is 7 years.
View full question & answer→Question 65 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$
AnswerThe given equation is $\sqrt2\text{x}^2+7\text{x}+5\sqrt2=0$.Comparing it with $ax^2 + bx + c = 0,$ we get
$\text{a}=\sqrt2,\ \text{b}=7$ and $\text{x}=5\sqrt2$
$\therefore$ Discriminant, $D = b^2 - 4ac$
$=(7)^2-4\times\sqrt2\times5\sqrt2$
$=49-40 = 9>0$
So, the given equation has real roots.
$\sqrt{\text{D}}=\sqrt9=3$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-7+3}{2\times\sqrt2}$
$=\frac{-4}{2\sqrt2}$
$=-\sqrt2$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-7-3}{2\times\sqrt2}$
$=\frac{-10}{2\sqrt2}$
$=-\frac{5\sqrt2}{2}$
Hence, $-\sqrt2$ and $=-\frac{5\sqrt2}{2}$ are the roots of the given equation.
View full question & answer→Question 75 Marks
A person on tour has Rs. 10800 for his expenses. If he extends his tour by 4 days, he has to cut down his daily expenses by Rs. 90. Find the original duration of the tour.
AnswerLet the original duration of a tour be x days.
Total expenditure on tour = Rs. 10800
$\therefore$ Expenditure per day $=\text{Rs. }\frac{10800}{\text{x}}$
Duration of extended tour = (x + 4)days
$\therefore$ Expenditure per day according to new schedule $=\text{Rs. }\frac{10800}{\text{x}+4}$
It is given that:
$\frac{10800}{\text{x}}-\frac{10800}{\text{x}+4}=90$
$\Rightarrow\frac{\text{10800x}+43200-\text{10800x}}{\text{x}^2+\text{4}}=90$
$\Rightarrow 43200 = 90x^2 + 360x$
$\Rightarrow 90x^2 + 360x - 43200 = 0$
$\Rightarrow x^2 + 4x - 480 = 0$
$\Rightarrow x^2 + 24x - 20x - 480 = 0$
$\Rightarrow x(x + 24) - 20(x + 24) = 0$
$\Rightarrow (x + 24)(x - 20) = 0$
$\Rightarrow x + 24 = 0 or x - 20 = 0$
$\Rightarrow x = -24 or x = 20$
Since number of days cannot be negative, $\text{x}\neq-24$
⇒ x = 20
Hence, the original duration of the tour is 20 days.
View full question & answer→Question 85 Marks
The area of a right-angled triangle is 96 sq metres. If the base is three times the altitude, find the base.
AnswerLet the altitude of triangle be x meter.
Hence, base = 3x meter
$\therefore$ Area of triangle $=\frac{1}{2}\times(3\text{x}\times\text{x})\text{cm}^2$
$=\frac{1}{2}\times\text{3x}^2=96$
$\Rightarrow\text{x}^2=\frac{96\times2}{3}$
$\Rightarrow\text{x}^2=64$
$\Rightarrow\text{x}=\sqrt{64}$
$\Rightarrow\text{x}=\pm8$
$\therefore\text{x}=8\ [\because$ lenght of altitude can never be negative$]$
Hence, altitude of triangle is 8cm.
And base of triangle = 3x = (3 × 8)cm = 24cm.
View full question & answer→Question 95 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$2\text{x}^2+6\sqrt3\text{x}-60=0$
AnswerThe given equation is: $2\text{x}^2+6\sqrt3\text{x}-60=0$ On comparing it with $ax^2 + bx + c = 0$, we get: $\text{x}=2,\ \text{b}=6\sqrt3$ and $\text{c}=-60$
$\therefore$ Discriminant D is given by:$ D = (b^2 - 4ac)$
$=\big(6\sqrt3\big)^2-4\times2\times(-60)$ $=108+480$
$=588>0$ So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{588}=14\sqrt3$
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-6\sqrt3+14\sqrt{3}}{2\times2}$ $=\frac{8\sqrt3}{4}$
$=2\sqrt3$ $\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-6\sqrt3-14\sqrt{3}}{2\times2}$ $=\frac{-20\sqrt3}{4}$
$=-5\sqrt3$Hence, $2\sqrt3$ and $-5\sqrt3$ are the roots of the given equation.
View full question & answer→Question 105 Marks
A takes 10 days than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.
AnswerSuppose B alone takes x days to finish the work
Then, A alone can finish it in (x - 10) days.
Now, (A's one day's work) + (B's one day work) $=\frac{1}{\text{x}-10}+\frac{1}{\text{x}}$
And, (A + B)'s one day's work $=\frac{1}{12}$
$\therefore\frac{1}{\text{x}-10}+\frac{1}{\text{x}}=\frac{1}{12}$
$\Rightarrow\frac{\text{x}+\text{x}-10}{\text{x}(\text{x}-10)}=\frac{1}{12}$
$\Rightarrow 12(2x - 10) = x(x - 10)$
$\Rightarrow 24x - 120 = x^2 - 10x$
$\Rightarrow x^2 - 34x + 120 = 0$
$\Rightarrow x^2 - 30x - 4x + 120 = 0$
$\Rightarrow x(x - 30) - 4(x - 30) = 0$
$\Rightarrow (x - 30)(x - 4) = 0$
$\Rightarrow x - 30 = 0 or x - 4 = 0$
$\Rightarrow x = 30 or x = 4$
Since number cannot be less that 10, $\text{x}\neq4$
$\Rightarrow x = 30$
Hence, B alone can finish the work in 30 days.
View full question & answer→Question 115 Marks
In a class test, the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained him in the two subjects separately.
AnswerLet the marks obtained by $P$ in mathematics $=x$ Then, marks obtained by him in science $=28-x$ It is given that $(x+\text { 3) }(28-x-4)=180$
$\Rightarrow(x+3)(24-x)=180$
$\Rightarrow 24 x-x^2+72-3 x=180$
$\Rightarrow 21 x-x^2=108$
$\Rightarrow x^2-12 x-9 x+108=0$
$\Rightarrow x(x-12)-9(x-12)=0$
$\Rightarrow(x-12)(x-9)=0$
$\Rightarrow x-12=0 \text { or } x-9=0$
$\Rightarrow x=12 \text { or } x=9 \text { When } x=12,28-x=28-12=16 \text { When } x=928-x=28-9=19 \text { Hence, }$
Marks in mathematics $=12$ and marks in science $=16$
or
Marks in mathematics $=9$ and marks in science $=19$
View full question & answer→Question 125 Marks
Solve the following quadratic equation:$x^2 - 4ax - b^2 + 4a^2 = 0$
Answer$x^2 - 4ax - b^2 + 4a^2 = 0\Rightarrow x^2 - 4ax + (4a^2 - b^2) = 0$
$\Rightarrow x^2 - 4ax + (2a + b)(2a - b) = 0$
$\Rightarrow x^2 - (2a + b)x - (2a - b)x + (2a + b)(2a - b) = 0$
$\Rightarrow x[x - (2a + b)] - (2a - b)[x + (2a + b)] = 0$
$\Rightarrow [x - (2a + b)][x - (2a - b)] = 0$
$\Rightarrow x - (2a + b) = 0 or x - (2a - b) = 0$
$\Rightarrow x = 2a + b or x = 2a - b$
View full question & answer→Question 135 Marks
The sum of the squares of two consecutive multiples of 7 is 1225. Find the multiples.
AnswerLet the required numbers be x and (x + 7).
Then, we have
$x^2+ (x + 7)^2 = 1225$
$\Rightarrow x^2 + x^2 + 14x + 49 = 1225$
$\Rightarrow 2x^2+ 14x - 1176 = 0$
$\Rightarrow x^2 + 7x - 588 = 0$
$\Rightarrow x^2 + 28x - 21x - 588 = 0$
$\Rightarrow x(x + 28) - 21(x + 28) = 0$
$\Rightarrow (x + 28)(x - 21) = 0$
$\Rightarrow x + 28 = 0 or x - 21 = 0$
$\Rightarrow x = -28 or x = 21$
$When x = -28$
$x + 7 = -28 + 7 = -21$
$When x = 21$
$x + 7 = 21 + 7 = 28$
Hence, the required numbers are 21, 28 or -21, -28.
View full question & answer→Question 145 Marks
The distance between Mumbai and Pune is 192km. Travelling by the Deccan Queen, it takes 48 minutes less than another train. Calculate the speed of the Deccan Queen if the speed of the two trains differ by 20km/hr.
AnswerLet the speed of the Deccan Queen = x kmph.
The speed of other train = (x - 20)kmph.
Then, time taken by Deccan Queen $=\Big(\frac{192}{\text{x}}\Big)\text{h}$
Time taken by other train $=\Big(\frac{192}{\text{x}-20}\Big)\text{h}$
Difference of time taken by two trains is $\frac{48}{60}\neq\frac{4}{5}\text{h}$
$\therefore\frac{192}{\text{x}-20}-\frac{192}{\text{x}}=\frac{4}{5}$
$\Rightarrow\frac{1}{\text{x}-20}-\frac{1}{\text{x}}=\frac{1}{240}$
$\Rightarrow\frac{\text{x}-\text{x}+20}{\text{x}^2-20\text{x}}=\frac{1}{240}$
$\Rightarrow x^2 - 20x - 4800 = 0$
$\Rightarrow x^2 - 80x + 60x - 4800 = 0$
$\Rightarrow x(x - 80) + 60(x - 80) = 0$
$\Rightarrow (x - 80)(x + 60) = 0$
$\Rightarrow x - 80 = 0 or x + 60 = 0$
$\Rightarrow x = -80 or x = -60$
Since the speed cannot be negative, $\text{x}\neq-60.$
$\Rightarrow x = 80$
Hence, the speed of Deccan Queen = 80km/h.
View full question & answer→Question 155 Marks
Solve the following quadratic equation:$abx^2 +(b^2 - ac)x - bc = 0$
Answer$abx^2 +(b^2 - ac)x - bc = 0\Rightarrow abx^2 + b^2x - acx - bc = 0$
$\Rightarrow bx(ax + b) - c(ax + b) = 0$
$\Rightarrow (ax + b)(bx + c) = 0$
$\Rightarrow (ax + b) = 0 or (bx - c) = 0$
$\Rightarrow\text{x}=\frac{-\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{c}}{\text{b}}$
Hence, $\frac{-\text{b}}{\text{a}}$ and $\frac{\text{c}}{\text{b}}$ are the roots of the given equation.
View full question & answer→Question 165 Marks
Solve the following equations by using the method of completing the square:
$\text{4x}^2+4\sqrt3+3=0$
Answer$\text{4x}^2+4\sqrt3+3=0$
$\Rightarrow\text{4x}^2+4\sqrt3\text{x}=-3$
$\Rightarrow(\text{2x})^2+2\times\text{2x}\times\sqrt3+\big(\sqrt3\big)^2\\=-3+\big(\sqrt3\big)^2$ $[$ Adding $\big(\sqrt3\big)^2$ on Both sides$]$
$\Rightarrow\big(\text{2x}+\sqrt3\big)^2=-3+3=0$
$\Rightarrow\text{2x}+\sqrt3=0$
$\Rightarrow\text{x}=-\frac{\sqrt3}{2}$
Hence, $-\frac{\sqrt3}2{}$ is the required root of the given equation.
View full question & answer→Question 175 Marks
A dealer sells an article for Rs. 75 and gains as much percent as the cost price of the article. Find the cost price of the article.
AnswerLet the cost price of an article be Rs. x. Then, gain = x% of x
⇒ Gain $=\text{Rs. }\Big(\text{x}\times\frac{\text{x}}{100}\Big)=\text{Rs}.\Big(\frac{\text{x}^2}{100}\Big)$
$\therefore$ S.P. = C.P. + Gain $=\text{x}+\frac{\text{x}^2}{100}$
But, S.P. = Rs. 75
$\Rightarrow\text{x}+\frac{\text{x}^2}{100}=75$
$\Rightarrow 100x + x^2 = 7500$
$\Rightarrow x^2 + 100x - 7500 = 0$
$\Rightarrow x^2 + 150x - 50x - 7500 = 0$
$\Rightarrow x(x + 150) - 50(x + 150) = 0$
$\Rightarrow (x + 150)(x - 50) = 0$
$\Rightarrow x + 150 = 0 or x - 50 = 0$
⇒ x = -150 or x = 50 Since the price cannot be negative,
$\text{x}\neq-150$ ⇒ x = 50
Thus, the cost price of an article is Rs. 50.
View full question & answer→Question 185 Marks
Find two consecutive multiples of 3 whose product is 648.
AnswerLet the required consecutive multiples of 3 be 3x and 3(x + 1).
Then, we have
$3x \times 3(x + 1) = 648$
$\Rightarrow 9x^2 + 9x - 648 = 0$
$\Rightarrow x^2 + x - 72 = 0$
$\Rightarrow x^2 + 9x - 8x - 72 = 0$
$\Rightarrow x(x + 9) - 8(x + 9) = 0$
$\Rightarrow (x + 9)(x - 8) = 0$
$\Rightarrow x + 9 = 0 or x - 8 = 0$
$\Rightarrow x = 9 or x = 8$
Since x is a positive integer, x ≠ -9
⇒ x = 8
⇒ 3x = 3 × 8 = 24 and 3(x + 1) = 3(9) = 27
Hence, the required consecutive multiples of 3 are 24 and 27.
View full question & answer→Question 195 Marks
The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third is 1 metre more than the shortest side, find the sides of the triangle.
AnswerLet the shorter side of triangle be x meter.
Then, its hypotenuse = (2x + 1) meter
And let the altitude = (x + 1) meter.
Then, $(2x - 1)^2 = x^2 + (x + 1)^2$
$\Rightarrow 4x^2 + 1 - 4x = x^2 + x^2 + 1 + 2x$
$\Rightarrow 2x^2- 6x = 0$
$\Rightarrow 2x(x - 3) = 0$
$\Rightarrow (x - 3) = 0 or 2x = 0$
$\Rightarrow x = 3 or x = 0$
⇒ x = 3 $[\because$ base cannot be zero$]$
thus, Base = 3m
Hypotenuse = (2 × 3 - 1)m = 5m
Altitude = (3 + 1)m = 4m.
View full question & answer→Question 205 Marks
The hypotenuse of a right-angled triangle is 20 metres. If the difference between the length of the other sides be 4 metres, find the other sides.
AnswerLet the other side of triangle be x and (x - 4) meters. By Pythagoras theorem, we have
$\Rightarrow x^2 + (x - 4)^2 = 330$
$\Rightarrow x^2 + x^2 + 16 - 8x = 400$
$\Rightarrow 2x^2 - 8x - 384 = 0$
$\Rightarrow x^2 - 4x - 192 = 0$
$\Rightarrow x^2 - 16x + 12x - 192 = 0$
$\Rightarrow x(x - 16) + 12(x - 16) = 0$
$\Rightarrow (x - 16)(x + 12) = 0$
$\Rightarrow x = 16 or x = -12$
$\Rightarrow x = 16$
$[\because$ height cannot be negative$]$
Thus, the height of the triangle be = 16m
And the base of the triangle = (16 - 4) = 12m.
View full question & answer→Question 215 Marks
Two pipes running together can fill a tank in $11\frac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.
AnswerSuppose the faster pipe takes x minutes to fill the tank.
Then, the slower pipe will take (x + 5) minutes to fill the tank.
$\therefore$ Protion of the tank filled by the faster pipe in one minute $=\frac{1}{\text{x}}$
⇒ Protion of the tank filled by the faster pipe in $\frac{100}{9}$ minutes.
$=\frac{1}{\text{x}}\times\frac{100}{9}$
$=\frac{100}{\text{9x}}$
Similarly, Protion of the tank filled by the slower pipe in $\frac{100}{9}$ minutes.
$=\frac{1}{\text{x}+5}\times\frac{100}{9}$
$=\frac{100}{9(\text{x}+5)}$
It is given that the tank is filled in $\frac{100}{9}$ minutes.
$\therefore\frac{100}{\text{9x}}+\frac{100}{9(\text{x}+5)}=1$
$\Rightarrow\frac{100}{\text{x}}+\frac{100}{\text{x}+5}=9$
$\Rightarrow\frac{100\text{x}+500+\text{100x}}{\text{x}^2+\text{5x}}=9$
$\Rightarrow 200x + 500 = 9x^2 + 45x$
$\Rightarrow 9x^2 - 155x - 500 = 0$
$\Rightarrow 9x^2 - 180x + 25x - 500 = 0$
$\Rightarrow 9x(x - 20) + 25(x - 20) = 0$
$\Rightarrow (x - 20)(9x + 25) = 0$
$\Rightarrow x - 20 = 0 or 9x + 25 = 0$
$\Rightarrow x = 20 $or $\text{x}=-\frac{25}{9}$
Since time cannot be negative, $\text{x}\neq-\frac{25}{9}$
⇒ x = 20
⇒ x + 5 = 20 + 5 = 25
Hence, the faster pipe fills the tank in 20 minutes and the slower pipe fills the tank in 25 minutes.
View full question & answer→Question 225 Marks
Divide 57 into two parts whose product is 680.
AnswerLet the one part be x.
Then, the other part will be (57 - x).
Thus, we have
$x(57 - x) = 680$
$\Rightarrow 57x - x^2= 680$
$\Rightarrow x^2 - 57x + 680 = 0$
$\Rightarrow x^2 - 17x - 40x + 680 = 0$
$\Rightarrow x(x - 17) - 40(x - 17) = 0$
$\Rightarrow (x - 17)(x - 40) = 0$
$\Rightarrow x - 17 = 0 or x - 40 = 0$
$\Rightarrow x = 17 or x = 40$
Hence, the two parts are 17 and 40.
View full question & answer→Question 235 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 2ax + (a^2 - b^2) = 0$
AnswerGiven, $x^2-2 a x+\left(a^2-b^2\right)=0$ On comparing it with $A x^2+b x+c=0$, we get: $A=1, B=-2 a$ and $C=\left(a^2-b^2\right)$ Discriminant $D$ is given by: $D=B^2-4 A C=(-2 a)^2-4 \times 1 \times\left(a^2-b^2\right)=4 a^2-4 a^2+4 b^2=4 b^2>0$ Hence, the roots of the equation are real.Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-2\text{a})+\sqrt{4\text{b}^2}}{2\times1}$
$=\frac{2\text{a}+2\text{b}}{2}$
$=\frac{2(\text{a}+\text{b})}{2}$
$=(\text{a}+\text{b})$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{2\text{a}-2\text{b}}{2}$
$=\frac{2(\text{a}-\text{b})}{2}$
$=(\text{a}-\text{b})$
Hence, the roots of the equation are (a + b) and (a - b).
View full question & answer→Question 245 Marks
Divide 27 into two parts such that the sum of their reciprocals is $\frac{3}{20}.$
AnswerLet the one part be $x$.
Then, the other part will be ( 27 - x).
Thus, we have
$\frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}$
$\Rightarrow \frac{27-x+x}{x(27-x)}=\frac{3}{20}$
$\Rightarrow \frac{27}{27 x-x^2}=\frac{3}{20}$
$\Rightarrow 27 \times 20=3(27 x)-3 x^2$
$\Rightarrow 540=81 x-3 x^2$
$\Rightarrow 3 x^2-81 x+540=0$
$\Rightarrow x^2-27 x+180=0$
$\Rightarrow x^2-15 x-12 x+180=0$
$\Rightarrow x(x-15)-12(x-15)=0$
$\Rightarrow(x-15)(x-12)=0$
$\Rightarrow x-15=0 \text { or } x-12=0$
$\Rightarrow x=15 \text { or } x=12$
Hence, the two parts are 15 and 12.
View full question & answer→Question 255 Marks
The sum of the squares of two consecutive positive integers is 365. Find the integers.
AnswerLet the required consecutive positive integers be x and (x + 1).
Then, we have
$x^2 + (x + 1)^2 = 365$
$\Rightarrow x^2 + x^2 + 2x + 1 = 365$
$\Rightarrow 2x^2 + 2x - 364 = 0$
$\Rightarrow x^2 + x + 182 = 0$
$\Rightarrow x^2 + 14x - 13x - 182 = 0$
$\Rightarrow x(x + 14) - 13(x + 14) = 0$
$\Rightarrow (x + 14)(x - 13) = 0$
$\Rightarrow x + 14 = 0 or x - 13 = 0$
$\Rightarrow x = -14 or x = 13$
Since x is a positive integer, $x \neq -14$
$\Rightarrow x = 13$
$\Rightarrow x + 1 = 13 + 1 = 14$
Hence, the required positive integers are 13 and 14.
View full question & answer→Question 265 Marks
Solve the following quadratic equation: $4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$
Answer$4 x^2-4 a^2 x+\left(a^4-b^4\right)=0 \Rightarrow 4 x^2-2\left(a^2+b^2\right) x-2\left(a^2-b^2\right) x+\left(a^4-b^4\right)=0$ $\Rightarrow 2 x\left[2 x-\left(a^2+b^2\right)\right]-\left(a^2-b^2\right)\left[2 x+\left(a^2+b^2\right)\right]=0$
$\Rightarrow\left[2 x -\left( a ^2+ b ^2\right)\right]\left[2 x -\left( a ^2- b ^2\right)\right]=0$
$\Rightarrow 2 x -\left( a ^2+ b ^2\right)=0$ or $2 x -\left( a ^2- b ^2\right)=0$
$\Rightarrow x =\frac{ a ^2+ b ^2}{2}$ or $x =\frac{ a ^2- b ^2}{2}$
View full question & answer→Question 275 Marks
If $a d \neq b c$ then prove that the equation $\left(a^2+b^2\right) x^2+2(a c+b d) x+\left(c^2+d^2\right)=0$ has no real roots.
AnswerCompare the given quadratic equation with $Ax ^2+ Bx + C =0$
Here $A=a^2+b^2, B=2(a c+b d)$ and $C=c^2+d^2$
Consider, $B^2-4 A C=[2(a c+b d)]^2-4\left(a^2+b^2\right) \times\left(c^2+d^2\right)$
$= 4[a^2c^2+ 2abcd + b^2d^2] - 4[a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2]$
$= 4a^2c^2 + 8abcd + 4b^2d^2 - 4a^2c^2 - 4a^2d^2 - 4b^2c^2 - 4b^2d^2$
$= 8abcd - 4a^2d^2 - 4b^2c^2$
$= -4[4a^2d^2 + 4bc^2 - 2abcd]$
$= -4[ad - bc]^2$
Hence the given equation has no real roots unless ad ≠ bc
View full question & answer→Question 285 Marks
The sum of two natural numbers is 28 and their product is 192. Find the numbers.
AnswerLet the required natural numbers be x and (28 - x).
Then, we have
$x \times (28 - x) = 192$
$\Rightarrow 28x - x^2 = 192$
$\Rightarrow x^2 - 28x + 192 = 0$
$\Rightarrow x^2 - 16x - 12x + 192 = 0$
$\Rightarrow x(x - 16) - 12(x - 16) = 0$
$\Rightarrow (x - 16)(x - 12) = 0$
$\Rightarrow x - 16 = 0 or x - 12 = 0$
$\Rightarrow x = 16 or x = 12$
Hence, the required numbers are 16 and 12.
View full question & answer→Question 295 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$2x^2 + ax - a^2 = 0$
AnswerThe given equation is: $2x^2 + ax - a^2 = 0$
Comparing it with $Ax^2 + Bx + C = 0$
$A = 2, B = a$ and $C = -a^2$
Discriminant, $D = B^2 - 4AC$
$= a^2 - 4 \times 2 \times -a^2$
$= a^2 + 8a^2$
$= 9a^2 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{9\text{a}^2}=\text{3a}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-\text{a}+\text{3a}}{2\times2}$
$=\frac{\text{2a}}{4}$
$=\frac{\text{a}}{2}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-\text{a}-\text{3a}}{2\times2}$
$=\frac{-\text{4a}}{4}$
$=-\text{a}$
View full question & answer→Question 305 Marks
In a class test, the sum of Kamal's marks in Mathematics and English is 40. Had he got 3 marks more in Mathematics and 4 marks less in English, the product of the marks would have been 360. Find his marks in two subjects separately.
AnswerLet the marks obtained by kamal in mathematice and english be $x$ and $y$.
$\therefore x+y=40 \ldots(1)$
$\text { and }(x+3)(y-4)=360$
From (1) $y=40-x$
Putting value of $y$ in (2)
$(x+3)(40-x-4)=360$
$\Rightarrow(x+3)(36-x)=360$
$\text { or } 36 x-x^2+108-3 x=360$
$\Rightarrow-x^2+33 x-252=0 \text { or } x^2-33 x+252=0$
$\Rightarrow x^2-21 x-12 x+252=0$
$\text { or } x(x-21)-12(x-21)=0$
$\Rightarrow(x-21)(x-12)=0$
$\therefore$ when $x-21=0, x=21$
when $x-12=0, x=12$
for $x=21,21+y=40 \therefore y=19$
$\text { for } x=12,12+y=40 \therefore y=28$
The marks obtained by kamal in mathematics and english respectively are $(21,19)$ or $(12,28)$.
View full question & answer→Question 315 Marks
A train travels 180km at a uniform speed. If the speed had been 9km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.
AnswerLet the uniform speed of the train be x km/hr.
Time taken to cover 180km $=\frac{180}{\text{x}}\ \text{hours}$
Time taken to cover 180km when the speed is increased by 9km/hr $=\frac{180}{\text{x}+9}\ \text{hours}$
$\therefore\frac{180}{\text{x}}-\frac{180}{\text{x}+9}=1$
$\Rightarrow\frac{\text{180x}+1620-\text{180x}}{\text{x}^2+\text{9x}}=1$
$\Rightarrow 1620 = x^2 + 9x$
$\Rightarrow x^2 + 9x - 1620 = 0$
$\Rightarrow x^2 + 45x - 36x - 1620 = 0$
$\Rightarrow x(x + 45) - 36(x + 45) = 0$
$\Rightarrow (x + 45)(x - 36) = 0$
$\Rightarrow x + 45 = 0 or x - 36 = 0$
$\Rightarrow x = -45 or x = 36$
Since the speed cannot be negative, $\text{x}\neq-45.$
⇒ x = 36
Hence, the uniform speed of the train is 36km/hr.
View full question & answer→Question 325 Marks
The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}.$ Find the fraction.
AnswerLet the denominator of the fraction be x.
Then, the numerator of the fraction will be (x - 3).
$\therefore$ Fraction $=\frac{\text{x}-3}{\text{x}}$
$\frac{\text{x}-3}{\text{x}+1}=\frac{\text{x}-3}{\text{x}}-\frac{1}{15}$
$\Rightarrow\frac{\text{x}-3}{\text{x}}-\frac{\text{x}-3}{\text{x}+1}=\frac{1}{15}$
$\Rightarrow\frac{(\text{x}+1)(\text{x}-3)-\text{x}(\text{x}-3)}{\text{x}(\text{x}+1)}=\frac{1}{15}$ when
$\Rightarrow\frac{(\text{x}^2-\text{2x}-3)-(\text{x}^2-\text{3x})}{\text{x}^2+\text{x}}=\frac{1}{15}$
$\Rightarrow\frac{\text{x}-3}{\text{x}^2+\text{x}}=\frac{1}{15}$
$\Rightarrow 15x - 45 = x^2 + x$
$\Rightarrow x^2 - 14x + 45 = 0$
$\Rightarrow x^2- 9x - 5x + 45 = 0$
$\Rightarrow x(x - 9) - 5(x - 9) = 0$
$\Rightarrow (x - 9)(x - 5) = 0$
$\Rightarrow x - 9 = 0 or x - 5 = 0$
$\Rightarrow x = 9 or x = 5$
When x = 9,
$x - 3 = 9 - 3 = 6$
$\Rightarrow$ Fraction $=\frac{6}{9}=\frac{2}{3}$
When x = 5,
x - 3 = 5 - 3 = 2
⇒ Fraction $=\frac{2}{5}$
Since, numerator is 3 less than the denominator, requierd fraction is $\frac{2}{5}.$
View full question & answer→Question 335 Marks
Find two consecutive positive odd integers whose product is 483.
AnswerLet the required consecutive positive odd integers be x and (x + 2).
Then, we have
$x \times (x + 2) = 483$
$\Rightarrow x^2 + 2x - 483 = 0$
$\Rightarrow x^2 + 23x - 21x - 483 = 0$
$\Rightarrow x(x + 23) - 21(x + 23) = 0$
$\Rightarrow (x + 23)(x - 21) = 0$
$\Rightarrow x + 23 = 0 or x - 21 = 0$
$\Rightarrow x = -23 or x = 21$
Since x is a positive integer, $x \neq -23$
$\Rightarrow x = 21$
$\Rightarrow x + 2 = 21 + 2 = 23$
Hence, the required consecutive positive odd integers are 21 and 23.
View full question & answer→Question 345 Marks
The sum of a natural number and its positive square root is 132. Find the number.
AnswerLet the natural number be $x$.
Then, its positive square root will be $\sqrt{ x }$
Accroding to given information, we have
$x+\sqrt{x}=132$
$\Rightarrow y^2+y=132 \text { where, } \sqrt{x}=y$
$\Rightarrow y^2+y-132=0$
$\Rightarrow y^2+12 y-11 y-132=0$
$\Rightarrow y(y+12)-11(y+12)=0$
$\Rightarrow(y+12)(y-11)=0$
$\Rightarrow y+12=0 \text { or } y-11=0$
$\Rightarrow y=-12 \text { or } y=11$
Since square root of a number cannot be negative, $y \neq-12$
Hence, $y=11$
$\Rightarrow \sqrt{x}=11$
$\Rightarrow x=121$
Thus, the required natural numbr is 121 .
View full question & answer→Question 355 Marks
The sum of a natural number and its square is 156. Find the number.
AnswerLet the natural number be x.
Then, its square will be $x^2$
Accroding to given information, we have
$\Rightarrow x + x^2 = 156$
$\Rightarrow x^2 + x - 156 = 0$
$\Rightarrow x^2 + 13x - 12x - 156 = 0$
$\Rightarrow x(x + 13) - 12(x + 13) = 0$
$\Rightarrow (x + 13)(x - 12) = 0$
$\Rightarrow x + 13 = 0 or x - 12 = 0$
$\Rightarrow x = -13 or x = 12$
Since x is a natural number, x ≠ -13
Hence, the required natural number is 12.
View full question & answer→Question 365 Marks
The length of a rectangle is twice its breadth and its area is 288cm$^2$. Find the dimensions of the rectangle.
AnswerLet the breadth of a rectangle = x cm
Then, lenght of the rectangle = 2x cm
$\therefore$ Area = lenght × breadth = 288cm$^2$
$\Rightarrow 2x \times x = 288$
$\Rightarrow 2x^2 = 288$
$\Rightarrow x^2 = 144$
$\Rightarrow\text{x}=\sqrt{144}$
$\Rightarrow\text{x}=\pm12$
$\Rightarrow\text{x}=12$ $[\because$ breadth cannot be negative$]$
Thus, breadth of rectangle = 12cm
And lenght of rectangle = (2 × 12) = 24cm.
View full question & answer→Question 375 Marks
A man is $3\frac{1}{2}$ times as old as his son. If the sum of the squares of their ages is 1325, find the ages of the father and the son.
Answer$\Rightarrow\text{x}^2+\Big(\frac{7}{2}\text{x}\Big)^2=1325$ $\Rightarrow\frac{\text{x}^2}{1}+\frac{\text{49x}^2}{4}=1325$ $\Rightarrow\frac{\text{4x}^2+\text{49x}^2}{4}=1325$ $\Rightarrow\text{53x}^2=1325\times4$ $\Rightarrow\text{x}^2=\frac{1325\times4}{53}=100$ $\Rightarrow\text{x}^2=\sqrt{100}=10$ Son's ages $\Rightarrow\text{x}=10$ Father's age$\Rightarrow\frac{7}{2}\times10=35\text{yrs.}$
View full question & answer→Question 385 Marks
A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.
AnswerLet the onglnal speed of the truck
= s km/hr New speed of the truck = (s + 20)km/hr Time taken for 150km + Time token for 200km = 5 $\frac{150}{\text{x}}+\frac{200}{(\text{s}+20)}=5$ $\Rightarrow\frac{150\text{x}+3000+\text{200s}}{\text{s}(\text{s}+20)}=5$ $\Rightarrow\frac{350\text{s}+3000}{\text{s}(\text{s}+20)}=5$
$\Rightarrow\frac{50(\text{7s}+60)}{\text{s}(\text{s}+20)}=5$
$10(7s + 60) = s(s + 20)$
$\Rightarrow 70s + 600 = s^2 + 20s$
$\Rightarrow s^2 - 50s - 600 = 0$
$\Rightarrow s^2 - 60s - 10s - 600 = 0$
$\Rightarrow s(s - 60) - 10(s - 60) = 0$
$\Rightarrow (s - 60)(s - 10) = 0$
$\Rightarrow s - 60 = 0 or s - 10 = 0$
$\Rightarrow s = 60 or s = 10$
⇒ s = 10 [Not possible]
$\therefore$ First speed of the truck = 60km/hr.
View full question & answer→Question 395 Marks
The area of a right-triangle is 600cm$^2$. If the base of the triangle exceeds the altitude by 10cm, find the dimensions of the triangle.
AnswerLet the altitude of triangle be x cm.Then, base of triangle is (x + 10)cm
$\therefore$ Area = 600cm$^2$ $\Rightarrow\frac{1}{2}\times$ base × altitude = $600cm^2$ $\Rightarrow\frac{1}{2}\times(\text{x}+10)\times\text{x}=600$$\Rightarrow x^2 + 10x = 1200$
$\Rightarrow x^2 + 10x - 1200 = 0$
$\Rightarrow x^2 + 40x - 30x - 1200 = 0$
$\Rightarrow x(x + 40) - 30(x + 40) = 0$
$\Rightarrow (x + 40)(x - 30) = 0$
$\Rightarrow x = -40 or x = 30$
$\therefore$ x = 30 $[\because$ length of altitude cannot be negative$]$Hence, altitude of triangle is 30cm and base of triangle 40cm.
$(Hypotenuse)^2 = (30)^2 + (40)^2$
$= 900 + 1600 = 2500$
$\therefore$ Hypotenuse = 50
View full question & answer→Question 405 Marks
The sum of the squares of two consecutive positive even numbers is 452. Find the numbers.
AnswerLet the required consecutive positive even numbers be x and (x + 2).
Then, we have
$x^2 + (x + 2)^2 = 452$
$\Rightarrow x^2 + x^2 + 4x + 4 = 452$
$\Rightarrow 2x^2 + 4x - 448 = 0$
$\Rightarrow x^2 + 2x - 224 = 0$
$\Rightarrow x^2 - 14x + 16x - 224 = 0$
$\Rightarrow x(x - 14) + 16(x - 14) = 0$
$\Rightarrow (x - 14)(x + 16) = 0$
$\Rightarrow x - 14 = 0 or x + 16 = 0$
$\Rightarrow x = 14 or x = -16$
Since x is a positive number, $x ≠ -16$
$\Rightarrow x = 14$
$\Rightarrow x + 2 = 14 + 2 = 16$
Hence, the required positive even numbers are 14 and 16.
View full question & answer→Question 415 Marks
A passenger train takes 2 hours less for a journey of 300km if its speed is increased by 5km/hr from its usual speed. Find its usual speed.
AnswerLet the usual speed of the passenger train be x km/hr.
Time taken to cover 300km $=\frac{300}{\text{x}}\ \text{hours}$
Time taken to cover 300km when the speed is increased by 5km/hr $=\frac{300}{(\text{x}+5)}\ \text{hours}$
It is given that the time taken to cover 300km is decreased by 2 hours.
$\therefore\frac{300}{\text{x}}-\frac{300}{\text{x}+5}=2$
$\Rightarrow\frac{300\text{x}+1500-300\text{x}}{\text{x}^2+5\text{x}}=2$
$\Rightarrow 1500 = 2x^2 + 10x$
$\Rightarrow 2x^2 + 10x - 1500 = 0$
$\Rightarrow x^2 + 5x - 750 = 0$
$\Rightarrow x^2 + 30x - 25x - 750 = 0$
$\Rightarrow x(x + 30) - 25(x + 30) = 0$
$\Rightarrow (x + 30)(x - 25) = 0$
$\Rightarrow x + 30 = 0 or x - 25 = 0$
$\Rightarrow x = -30 or x = 25$
Since the speed cannot be negative, $\text{x}\neq30.$
⇒ x = 25
Hence the original speed of train is 25kmph.
View full question & answer→Question 425 Marks
The product of two consecutive positive integers is 306. Find the integers.
AnswerLet the required consecutive positive integers be x and (x + 1).
Then, we have
$x \times (x + 1) = 306$
$\Rightarrow x^2 + x - 306 = 0$
$\Rightarrow x^2 + 18x - 17x - 306 = 0$
$\Rightarrow x(x + 18) - 17(x + 18) = 0$
$\Rightarrow (x + 18)(x - 17) = 0$
$\Rightarrow x + 18 = 0 or x - 17 = 0$
$\Rightarrow x = -18 or x = 17$
Since x is a positive integer, x ≠ -18
⇒ x = 17
⇒ x + 1 = 17 + 1 = 18
Hence, the required positive integers are 17 and 18.
View full question & answer→Question 435 Marks
300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students.
AnswerLet the number of students be x, then
$\frac{300}{\text{x}}-\frac{300}{(\text{x}+10)}=1$
$\Rightarrow\frac{1}{\text{x}}-\frac{1}{(\text{x}+10)}=\frac{1}{300}$
$\Rightarrow\frac{\text{x}+10-\text{x}}{\text{x}(\text{x}+10)}=\frac{1}{300}$
$\Rightarrow x(x + 10) = 3000$
$\Rightarrow x^2+ 10x - 3000 = 0$
$\Rightarrow x^2 + 60x - 50x - 3000 = 0$
$\Rightarrow x(x + 60) - 50(x + 60) = 0$
$\Rightarrow (x + 60)(x - 50) = 0$
$\Rightarrow x + 60 = 0 or x - 50 = 0$
$\Rightarrow x = -60 or x = 50$
⇒ x = 50 $(\because$ number of students cannot be negative$)$
Hence the number of students is 50.
View full question & answer→Question 445 Marks
Solve the following equations by using the method of completing the square:
$\text{x}^2-\big(\sqrt2+1\big)\text{x}+\sqrt2=0$
Answer$\text{x}^2-\big(\sqrt2+1\big)\text{x}+\sqrt2=0$
$\Rightarrow\text{x}^2-\big(\sqrt2+1\big)\text{x}=-\sqrt2$
$\Rightarrow\text{x}^2-2\times\text{x}\times\Big(\frac{\sqrt2+1}{2}\Big)+\Big(\frac{\sqrt2+1}{2}\Big)^2\\=-\sqrt2+\Big(\frac{\sqrt2+1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{\sqrt2+1}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big[\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)\Big]^2=\frac{-4\sqrt2+2+1+2\sqrt2}{4}$
$=\frac{2-2\sqrt2+1}{4}=\Big(\frac{\sqrt2-1}{2}\Big)^2$
$\Rightarrow\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=\pm\Big(\frac{\sqrt2-1}{2}\Big)$ (Taking square root on both sides)
$\Rightarrow\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=\Big(\frac{\sqrt2-1}{2}\Big)$ or $\text{x}-\Big(\frac{\sqrt2+1}{2}\Big)=-\Big(\frac{\sqrt2-1}{2}\Big)$
$\Rightarrow\text{x}=\Big(\frac{\sqrt2+1}{2}\Big)+\Big(\frac{\sqrt2-1}{2}\Big)$ or $\text{x}=\Big(\frac{\sqrt2+1}{2}\Big)-\Big(\frac{\sqrt2-1}{2}\Big)$
$\Rightarrow\text{x}=\frac{2\sqrt2}{2}=\sqrt2$ or $\text{x}=\frac{2}2{}=1$
Hence, $\sqrt2$ and 1 are the roots of the given equation.
View full question & answer→Question 455 Marks
A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.
AnswerLet the tens digit and units digit of required number be x and y respectively.
$\therefore\ \text{xy}=14$
$\Rightarrow\text{y}=\frac{14}{\text{x}}\ \dots(1)$
the number = (10x + y)
(10x + y) + 45 = (10x + y)
9x - 9y = -45
x - y = -5 ...(2)
Putting $\text{y}=\frac{14}{\text{x}}$ from (1) in (2), we get
$\text{x}-\frac{14}{\text{x}}=-5$
$\Rightarrow x^2 + 5x - 14 = 0$
$\Rightarrow x^2 + 7x - 2x - 14 = 0$
$\Rightarrow x(x + 7) - 2(x + 7) = 0$
$\Rightarrow (x + 7)(x - 2) = 0$
$\Rightarrow x + 7 = 0 or x - 2 = 0$
$\Rightarrow x = - 7 or x = 2$
$\therefore$ x = 2 $[\because$ a digit cannot be negative$]$
Putting x = 2 in (1), we get y = 8
The ten digit is 2 and unit digit is 7
Hence, the required number is 27.
View full question & answer→Question 465 Marks
If the price of a book is reduced by Rs. 5, a person can buy 4 more books for Rs. 600. Find the original price of the book.
AnswerLet the original price of the book = Rs. x.
$\therefore$ Number of books bought for Rs. 600 $=\frac{600}{\text{x}}$
Reduced price of the book = Rs. (x - 5)
$\therefore$ Number of books bought for Rs. 600 $=\frac{600}{\text{x}-5}$
It is given that:
$\frac{600}{\text{x}-5}-\frac{600}{\text{x}}=4$
$\Rightarrow\frac{\text{600x}-\text{600x}+3000}{\text{x}^2-\text{5x}}=4$
$\Rightarrow 3000 = 4x^2 - 20x$
$\Rightarrow 4x^2 - 20x - 3000 = 0$
$\Rightarrow x^2 - 5x - 750 = 0$
$\Rightarrow x^2 - 30x + 25x - 750 = 0$
$\Rightarrow x(x - 30) + 25(x - 30) = 0$
$\Rightarrow (x - 30)(x + 25) = 0$
$\Rightarrow x - 30 = 0 or x + 25 = 0$
$\Rightarrow x = 30 or x = -25$
Since price of the book cannot be negative, $\text{x}\neq-25$
⇒ x = 30
Hence, the original price of a book is Rs. 30
View full question & answer→Question 475 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\frac{\text{m}}{\text{n}}\text{x}^2+\frac{\text{n}}{\text{m}}=1-2\text{x}$
AnswerThe given equation is:$\frac{\text{m}}{\text{n}}\text{x}^2+\frac{\text{n}}{\text{m}}=1-2\text{x}$
$\Rightarrow\frac{\text{m}^2\text{x}^2+\text{n}^2}{\text{mn}}=1-2\text{x}$
$\Rightarrow\text{m}^2\text{x}^2+\text{n}^2=\text{mn}-2\text{mnx}$
$\Rightarrow\text{m}^2\text{x}^2+2\text{mnx}+\text{n}^2-\text{mn=0}$
This equation is the form $ax^2+ bx + c = 0, where a = m^2, b = 2mn and c = n^2 - mn$.$\therefore$ Discriminant $\text{D}=\text{b}^2-4\text{ac}$
$=(2\text{mn})^2-4\times\text{m}^2\times(\text{n}^2-\text{mn})$
$=4\text{m}^2\text{n}^2-4\text{m}^2\text{n}^2+4\text{m}^3\text{n}$
$=4\text{m}^3\text{n}>0$
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{4\text{m}^3\text{n}}=2\text{m}\sqrt{\text{mn}}$
$\therefore\alpha =\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-2\text{mn}+2\text{m}\sqrt{\text{mn}}}{2\times\text{m}^2}$
$=\frac{2\text{m}\big(-\text{n}+\sqrt{\text{mn}}\big)}{2\text{m}^2}$
$=\frac{-\text{n}+\sqrt{\text{mn}}}{\text{m}}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-2\text{mn}-2\text{m}\sqrt{\text{mn}}}{2\times\text{m}^2}$
$=\frac{-2\text{m}\big(\text{n}+\sqrt{\text{mn}}\big)}{2\text{m}^2}$
$=\frac{-\text{n}+\sqrt{\text{mn}}}{\text{m}}$
Hence, $\frac{-\text{n}+\sqrt{\text{mn}}}{\text{m}}$ and $\frac{-\text{n}-\sqrt{\text{mn}}}{\text{m}}$ are the roots of the given equation.
View full question & answer→Question 485 Marks
The difference of two natural numbers is 3 and the difference of their reciprocals is $\frac{3}{28}.$ Find the numbers.
AnswerLet the required numbers be x and (x - 3).
Then, we have
x > x - 3
$\Rightarrow\frac{1}{\text{x}}<\frac{1}{\text{x}-3}$
$\therefore\frac{1}{\text{x}-3}-\frac{1}{\text{x}}=\frac{3}{28}$
$\Rightarrow\frac{\text{x}-\text{x}+3}{\text{x}(\text{x}-3)}=\frac{3}{28}$
$\Rightarrow 84 = 3x^2 - 9x$
$\Rightarrow 3x^2 - 9x - 84 = 0$
$\Rightarrow x^2 - 3x - 28 = 0$
$\Rightarrow x^2 - 7x + 4x - 28 = 0$
$\Rightarrow x(x - 7) + 4(x - 7) = 0$
$\Rightarrow (x - 7)(x + 4) = 0$
$\Rightarrow x - 7 = 0 or x + 4 = 0$
$\Rightarrow x = 7 or x = -4$
Since x is a natural number, x ≠ -4
⇒ x = 7 and x - 3 = 4
Hence, the required numbers are 7 and 4.
View full question & answer→Question 495 Marks
Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers.
AnswerLet the three consecutive numbers be x, x + 1, x + 2.
Sum of square of first and product of the other two
$\Rightarrow x^2 + (x + 1)(x + 2) = 46$
$\Rightarrow x^2 + (x^2 + 3x + 2) = 46 or 2x^2 + 3x - 44 = 0$
$\Rightarrow 2x^2 + 11x - 8x - 44 = 0 or x(2x + 11) - 4(2x - 11) = 0$
$\Rightarrow (x - 4)(2x + 11) = 0$
$\therefore\ \text{x}=4$ or $\frac{-11}{2}$
But $\text{x}\neq\frac{-11}{2}\ \ \therefore\text{x}=4$
$\therefore$ required numbers are 4, 5 and 6
View full question & answer→Question 505 Marks
By using the method of completing the square, show that the equation $2x^2 + x + 4 = 0$ has no real roots.
Answer$2x^2 + x + 4 = 0$
$\Rightarrow 4x^2 + 2x + 8 = 0 (Multiplying both sides by 2)$
$\Rightarrow 4x^2 + 2x = -8$
$\Rightarrow(\text{2x})^2+2\times\text{2x}\times\frac{1}{2}+\Big(\frac{1}{2}\Big)^2\\=-8+\Big(\frac{1}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{1}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}+\frac{1}{2}\Big)^2=-8+\frac{1}{4}$
$=-\frac{31}{4}<0$
But $\Big(\text{2x}+\frac{1}{2}\Big)^2$ cannot be negative for any real value of x.
So, there is no real value of x satisfying the given equation.
Hence, the given equation has no real roots.
View full question & answer→Question 515 Marks
Find the value of a and b for which $\text{x}=-\frac{3}4{}$ and x = -2 are the root of the equation $ax^2 + bx - 6 = 0.$
AnswerMultiplying (2) by 4 adding the result from (1), we get
11a = 44
⇒ a = 4
Putting a = 4 in (1), we get
3 × 4 + 4b = 32
⇒ 4b = 32 - 12 = 20
$\Rightarrow\text{b}=\frac{20}{4}=5$
$\therefore$ a = 4 and b = 5
View full question & answer→Question 525 Marks
Some students planned a picnic. The total budget for food was Rs. 2000. But, 5 students failed to attend the picnic and thus the cost for food for each member increased by Rs. 20. How many students attended the picnic and how much did each student pay for the food?
AnswerLet the number of students be x.
Then, cost of food for each student $=\text{Rs. }\frac{2000}{\text{x}}$
If the number of students decreased by 5, then
New cost of food for each students $=\text{Rs. }\frac{2000}{\text{x}-5}$
It is given that:
$\frac{2000}{\text{x}-5}-\frac{2000}{\text{x}}=20$
$\Rightarrow\frac{\text{2000x}-\text{2000x}+10000}{\text{x}^2-\text{5x}}=20$
$\Rightarrow 10000 = 20x^2 - 100x$
$\Rightarrow 20x^2 - 100x - 10000 = 0$
$\Rightarrow x^2 - 5x - 500 = 0$
$\Rightarrow x^2 - 25x + 20x - 500 = 0$
$\Rightarrow x(x - 25) + 20(x - 25) = 0$
$\Rightarrow (x - 25)(x + 20) = 0$
$\Rightarrow x - 25 = 0 or x + 20 = 0$
$\Rightarrow x = 25 or x = -20$
Since number of students cannot be negative, $\text{x}\neq-20$
⇒ x = 25
⇒ x - 5 = 20
$\Rightarrow\frac{2000}{\text{x}-5}=\frac{2000}{20}=100$
Hence, the number of students who attended the picnic is 20 and the cost of food dor each student is Rs. 100.
View full question & answer→Question 535 Marks
Two natural numbers differ by 3 and their product is 504. Find the numbers.
AnswerLet the required number be x and (x - 3).
Then, we have
$x(x - 3) = 504$
$\Rightarrow x^2 - 3x - 504 = 0$
$\Rightarrow\text{x}=\frac{3\pm\sqrt{9+2016}}{2}$
$=\frac{3\pm\sqrt{2025}}{2}$
$=\frac{(3\pm45)}{2}$
$\Rightarrow\text{x}=\frac{3+45}{2}=24$ or $\text{x}=\frac{(3-45)}{2}=\frac{-42}{2}=-21$
Hence, the required numbers are (24, 21) or (-21 and - 24).
View full question & answer→Question 545 Marks
The speed of a boat in still water is 15km/hr. It goes 30km upstream and return back at the same point in 4 hours 30 minutes. Find the speed of the stream.
AnswerLet the speed of the stream be x km/h.It is given that the speed of a boat in still water is 15km/h.
Now,
Speed of the boat upstream = Speed of the boat in still water - Speed of the stream = (15 - x)km/h
Speed of the boat downstream = Speed of the boat in still water + Speed of the stream = (15 + x)km/h
We know that,
$\text{Time}=\frac{\text{Distance}}{\text{Speed}}$
Time is taken for upstream journey + Time taken for the downstream journey = 4h 30min
$\therefore\frac{30}{15-\text{x}}+\frac{30}{15+\text{x}}=4\frac{1}{2}$
$\Rightarrow30\Big[\frac{(15+\text{x})+(15-\text{x})}{(15+\text{x})(15-\text{x})}\Big]=\frac{9}{2}$
$\Rightarrow30\Big(\frac{30}{225-\text{x}^2}\Big)=\frac{9}{2}$
$\Rightarrow225-\text{x}^2=\frac{30\times30\times2}{9}$
$\Rightarrow225-\text{x}^2=200$
$\Rightarrow\text{x}^2=225-200=25$
$\Rightarrow\text{x}=\pm5$
Since speed cannot be negative, therefore, x = 5.
Thus, the speed of the stream is 5km/h.
View full question & answer→Question 555 Marks
Two taps running together can fill a tank in $3\frac{1}{13}\ \text{hours.}$ If one pipe takes 3 hours more than the other to fill the tank then how much time will each tap take to fill the tank
AnswerLet the faster tap takes x minutes to fill the tank.
Then, the other tap takes (x + 3) minute
$\therefore\frac{1}{\text{x}}+\frac{1}{(\text{x}+3)}=\frac{13}{40}$
$\Rightarrow\frac{(\text{x}+3)+\text{x}}{\text{x}(\text{x}+3)}=\frac{13}{40}$
$\Rightarrow 40(2x + 3) = 13(x^2 + 3x)$
$\Rightarrow 13x^2 - 41x - 120 = 0$
$\Rightarrow 13x^2 - 65x + 24x - 120 = 0$
$\Rightarrow 13x(x - 5) + 24(x - 5) = 0$
$\Rightarrow (13x + 24)(x - 5) = 0$
$\Rightarrow x - 5 = 0 or 13x + 24 = 0$
⇒ x = 5 or $\text{x}=-\frac{24}{3}$
Since time cannot be negative, $\text{x}\neq-\frac{24}{3}$
⇒ x = 5
The faster tap takes 5 minutes to fill the tank.
Then, the other tap takes (5 + 3) minutes = 8 minutes.
View full question & answer→Question 565 Marks
The difference of two natural numbers is 5 and the difference of their reciprocals is $\frac{5}{14}.$ Find the numbers.
AnswerLet the required numbers be x and (x - 5).
Then, we have
x > x - 5
$\Rightarrow\frac{1}{\text{x}}<\frac{1}{\text{x}-5}$
$\therefore\frac{1}{\text{x}-5}-\frac{1}{\text{x}}=\frac{5}{14}$
$\Rightarrow\frac{\text{x}-\text{x}+5}{\text{x}(\text{x}-5)}=\frac{5}{14}$
$\Rightarrow 70 = 5x^2 - 25x$
$\Rightarrow 5x^2 - 25x - 70 = 0$
$\Rightarrow x^2 - 5x - 14 = 0$
$\Rightarrow x^2 - 7x + 2x - 14 = 0$
$\Rightarrow x(x - 7) + 2(x - 7) = 0$
$\Rightarrow (x - 7)(x + 2) = 0$
$\Rightarrow x - 7 = 0 or x + 2 = 0$
$\Rightarrow x = 7 or x = -2$
Since x is a natural number, x ≠ -2
⇒ x = 7 and x - 5 = 2
Hence, the required numbers are 7 and 2.
View full question & answer→Question 575 Marks
If the roots of the equations $ax^2 + 2bx + c = 0$ and $\text{bx}^2-2\sqrt{\text{ac}}\text{x}+\text{b}=0$ are simultaneously real then prove that $b^2 = ac.$
AnswerIt is given that the roots of the equation $ax^2 + 2bx + c = 0 $are real.
$\therefore\text{D}_1=(\text{2b})^2-4\times\text{a}\times\text{c}\ge0$
$\Rightarrow4(\text{b}^2-\text{ac})\ge0$
$\Rightarrow\text{b}^2-\text{ac}\ge0\ \dots(1)$
Also, the roots of the equation $\text{bx}^2-2\sqrt{\text{ac}}\text{x}+\text{b}=0$ are real.
$\therefore\text{D}_2=\big(-2\sqrt{\text{ac}}\big)^2-4\times\text{b}\times\text{b}\ge0$
$\Rightarrow4(\text{ac}-\text{b}^2)\ge0$
$\Rightarrow-4(\text{b}^2-\text{ac})\ge0$
$\Rightarrow\text{b}^2-\text{ac}\le0\ \dots(2)$
The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if:
$b^2- ac = 0$
$\Rightarrow b^2 = ac$
View full question & answer→Question 585 Marks
A train covers a distance of 480km at a uniform speed. If the speed had been 8km/h less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.
AnswerLet the usual speed of the train be x km/hr
Time taken to cover 480km $=\frac{480}{\text{x}}\ \text{hours.}$
Time taken to cover 480km when the speed is decreased by 8km/hr $=\frac{480}{\text{x}-8}\ \text{hours}.$
$\therefore\frac{480}{\text{x}}=\frac{480}{\text{x}-8}-3$
$\Rightarrow\frac{1500}{\text{x}}-\frac{1500}{\text{x}+100}=\frac{1}{2}$
$\Rightarrow\frac{480\text{x}-3840-480\text{x}}{\text{x}^2-\text{8x}}=-3$
$\Rightarrow -3840 = -3x^2 + 24x$
$\Rightarrow 3x^2 - 24x - 3840 = 0$
$\Rightarrow x^2 - 8x - 1280 = 0$
$\Rightarrow x^2 - 40x + 32x - 1280 = 0$
$\Rightarrow x(x - 40) + 32(x - 40) = 0$
$\Rightarrow (x - 40)(x + 32) = 0$
$\Rightarrow x - 40 = 0 or x + 32 = 0$
$\Rightarrow x = 40 or x = -32$
Since speed cannot be negative, $\text{x}\neq-32.$
⇒ x = 40
Hence, the usual speed of the plane is 40km/hr.
View full question & answer→Question 595 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$15x^2 - 28 = x$
AnswerGiven,
$15x^2 - 28 = x$
$15x^2 - x - 28 = 0$
$On comparing it with ax^2 + bx + c = 0, we get:$
$a = 15, b = -1 and c = -28$
Discriminant D is given by:
$D = (b^2- 4ac)$
$= (-1)^2 - 4 \times 15 \times (-28)$
$= 1 - (-1680)$
$= 1 + 1680$
$= 1681$
$= 1681 > 0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-1)+\sqrt{1681}}{2\times15}$
$=\frac{1+41}{30}$
$=\frac{42}{30}$
$=\frac{7}{5}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-1)-\sqrt{1681}}{2\times15}$
$=\frac{1-41}{30}$
$=\frac{-40}{30}$
$=\frac{-4}{3}$
Thus, the roots of the equation are $\frac{7}{5}$ and $\frac{-4}{3}.$
View full question & answer→Question 605 Marks
The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number. Find the numbers.
AnswerLet x, y be the two natural numbers and x > y.
Then,
$\therefore$ $x^2 - y^2 = 45 ...(1)$
Also, square of smaller number = 4 larger number
$\Rightarrow y^2 = 4x ...(2)$
Putting value of $y^2$ from (1), we get
$x^2 - 4x = 45$
$\Rightarrow x^2 - 4x - 45 = 0$
$\Rightarrow x^2 - 9x + 5x - 45 = 0 or x(x - 9) + 5(x - 9) = 0$
$\Rightarrow (x - 9)(x - 5) = 0$
$\Rightarrow x = 9, -5$
But $\text{x}\neq-5\ \ \therefore\text{x}=9$
From (2), $y^2 = 4x = 4 \times 9 = 36$
$\therefore$ y = 6
Thus, the two required numbers are 9 and 6
View full question & answer→Question 615 Marks
A motor boat whose speed in still water is 18 km/hr, takes 1 hour more to go 24 km upstream than to return to the same spot. Find the speed of the stream.
AnswerLet the speed of the stream be x km/h. The speed of boat in still stream = 18km/h. Speed of boat up to the stream = 18 - x km/h $\therefore$ Time taken by boat to go up the stream 24km $=\frac{24}{18-\text{x}}\text{h}$
$\therefore$ Time taken by boat to go down the stream
$=\frac{24}{18+\text{x}}\text{h}$
Time taken by the boat to go up the stream is 1 hour more than the time taken down the stream.
$\therefore\frac{24}{18-\text{x}}-\frac{24}{18+\text{x}}=1$
$\Rightarrow\frac{1}{18-\text{x}}-\frac{1}{18+\text{x}}=\frac{1}{24}$
$\Rightarrow\frac{18+\text{x}-(18-\text{x})}{(18+\text{x})(18-\text{x})}=\frac{1}{24}$
$\Rightarrow\frac{2\text{x}}{324-\text{x}^2}=\frac{1}{24}$
$\Rightarrow 324 - x^2 = 48x$
$\Rightarrow x^2 + 48x - 324 = 0$
$\Rightarrow x^2 + 54x - 6x - 324 = 0$
$\Rightarrow x(x + 54) - 6(x + 54) = 0$
$\Rightarrow (x + 54)(x - 6) = 0$
$\Rightarrow x + 54 = 0 or x - 6 = 0$
$\Rightarrow x = -54 or x = 6$
Since the speed cannot be negative,
$\text{x}\neq-54.$
$\therefore$ x = 6
Hence, the speed of the stream = 6km/h.
View full question & answer→Question 625 Marks
The sum of two natural numbers is 15 and the sum of their reciprocals is $\frac{3}{10}.$ Find the numbers.
AnswerLet the required numbers be x and (15 - x).
Then, we have
$\frac{1}{\text{x}}+\frac{1}{15-\text{x}}=\frac{3}{10}$
$\Rightarrow\frac{15-\text{x}+\text{x}}{\text{x}(15-\text{x})}=\frac{3}{10}$
$\Rightarrow 150 = 45x - 3x^2$
$\Rightarrow 3x^2 - 45x + 150 = 0$
$\Rightarrow x^2 - 15x + 50 = 0$
$\Rightarrow x^2 - 10x - 5x + 50 = 0$
$\Rightarrow x(x - 10) - 5(x - 10) = 0$
$\Rightarrow (x - 10)(x - 5) = 0$
$\Rightarrow x - 10 = 0 or x - 5 = 0$
$\Rightarrow x = 10 or x = 5$
Hence, the required numbers are 5 and 10.
View full question & answer→Question 635 Marks
A man busy a number of pens for Rs. 180. If he had bought 3 more pens for the same amount, each pen would have cost him Rs. 3 less. How many pens did he buy?
AnswerLet the number of pens bought be x.
Then, cost of x pens = Rs. 180
⇒ cost of pen pen $=\text{Rs. }\frac{180}{\text{x}}$
If number of pens bought is (x + 3), then
Cost of one pen $=\text{Rs. }\frac{180}{\text{x}+3}$
It is given that
$\frac{180}{\text{x}}-\frac{180}{\text{x}+3}=3$
$\Rightarrow\frac{\text{180x}+\text{540x}-\text{180x}}{\text{x}^2+\text{3x}}=3$
$\Rightarrow 540x = 3x^2 + 9x$
$\Rightarrow 3x^2 + 9x - 540 = 0$
$\Rightarrow x^2 + 9x - 108 = 0$
$\Rightarrow x^2 + 15x - 12x - 108 = 0$
$\Rightarrow x(x + 15) - 12(x + 15) = 0$
$\Rightarrow (x + 15)(x - 12) = 0$
$\Rightarrow x + 15 = 0 or x - 12 = 0$
$\Rightarrow x = -15 or x = -12$
Since number of pens cannot be negative, $\text{x}\neq-15$
⇒ x = 12
Hence, he bought 12 pens.
View full question & answer→Question 645 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$2\text{x}^2-2\sqrt2\text{x}+1=0$
AnswerThe given equation is $2\text{x}^2-2\sqrt2\text{x}+1=0$.Comparing it with $ax^2 + bx + c = 0,$ we get
$\text{a}=2,\ \text{b}=-2\sqrt2$ and c = 1
$\therefore$ Discriminant, $D = b^2 - 4ac$
$=(-2\sqrt2)^2-4\times2\times1$
$=8-8=0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=0$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(-2\sqrt2\big)+\sqrt{0}}{2\times2}$
$=\frac{2\sqrt2}{4}$
$=\frac{\sqrt2}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big(-2\sqrt2\big)-\sqrt0}{2\times2}$
$=\frac{2\sqrt2}{4}$
$=\frac{\sqrt2}{2}$
Hence, $\frac{\sqrt2}{2}$is the repeated root of the given equation.
View full question & answer→Question 655 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$x^2 - kx + 9 = 0$
AnswerThe given equation is $x^2-k x+9=0$
$\therefore D=(-k)^2-4 \times 1 \times 9$
$D=k^2-36$
The given equation has real and distinct roots if $D>0$.
$\therefore k^2-36>0$
$\Rightarrow(k-6)(k+6)>0$
$\Rightarrow k<-6 \text { or } k>6$
View full question & answer→Question 665 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$x^2 + x + 2 = 0$
AnswerThe given equation is $x ^2+ x +2=0$ Comparing it with $ax ^2+ bx + c =0$, we get
$a=1, b=1 \text { and } c=2$
$\therefore \text { Discriminant, } D=b^2-4 a c$
$=1^2-4 \times 1 \times 2$
$=1-8$
$=-7<0$
Hence, the given equation has no real roots (or real roots does not exist).
View full question & answer→Question 675 Marks
Solve the following equations by using the method of completing the square:
$x^2 + 8x - 2 = 0$
Answer$x^2 + 8x - 2 = 0$
$\Rightarrow x^2 + 8x = 2$
$\Rightarrow x^2 + 2 \times x \times 4 + 4^2 = 2 + 4^2 (Adding 4^2 on both sides)$
$\Rightarrow (x + 4)^2= 2 + 16 = 18$
$\Rightarrow\text{x}+4=\pm\sqrt{18}=\pm3\sqrt2$ (Taking square root on both sides)
$\Rightarrow\text{x}+4=3\sqrt2$ or $\text{x}+4=-3\sqrt2$
$\Rightarrow\text{x}=-4+3\sqrt2$ or $\text{x}=-4-3\sqrt2$
Hence, $\big(-4+3\sqrt2\big)$ and $\big(-4-3\sqrt2\big)$ are the roots of the given equation.
View full question & answer→Question 685 Marks
The sum of the ages of a boy and his brother is 25 years, and the product of their ages in years is 126. Find their ages.
AnswerLet the age of a boy be x years. Then,His brother's age = (25 - x) years
It is given that
$x(25 - x) = 126$
$\Rightarrow 25x - x^2 = 126$
$\Rightarrow x^2 - 25x + 126$
$\Rightarrow x^2 - 18x - 7x + 126 = 0$
$\Rightarrow x(x - 18) - 7(x - 18) = 0$
$\Rightarrow (x - 18)(x - 7) = 0$
$\Rightarrow x - 18 = 0 or x - 7 = 0$
$\Rightarrow x = 18 or x = 7$
Hence, their ages are 18 years and 7 years.
View full question & answer→Question 695 Marks
Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages.
AnswerSuppose two years ago, son's age be $x$ years. Then, man's age two years ago $=3 x^2$ years Present age of son $=(x+2)$ years And, present age of father $=\left(3 x^2+2\right)$ years Three year's hence, we have Son's age $=(x+2+3)=(x+5)$ years Father's age $=\left(3 x^2+2+3\right)=\left(3 x^2+5\right)$ years It is given that
$3x^2 + 5 = 4(x + 5)$
$\Rightarrow 3x^2 + 5 = 4x + 20$
$\Rightarrow 3x^2 - 4x - 15 = 0$
$\Rightarrow 3x^2 - 9x + 5x - 15 = 0$
$\Rightarrow 3x(x - 3) + 5(x - 3) = 0$
$\Rightarrow (x - 3)(3x + 5) = 0$
$\Rightarrow x - 3 = 0 or 3x + 5 = 0$
⇒ x = 3 or $\text{x}=\frac{-5}{3}$ Since, age cannot be in fraction,
$\text{x}\neq\frac{-5}{3}.$
$\Rightarrow x = 3$
Thus, present age of son $=(x+2)=5$ years.
And, present age of father $=\left(3 x^2+2\right)=3(3)^2+2=29$ years.
View full question & answer→Question 705 Marks
The area of a right-angled triangle is 165 sq metres. Determine its base and altitude if the latter exceeds the former by 7 metres.
AnswerLet the base of triangle be x meter. Then, altitude of triangle = (x + 7)meter. $\therefore$ Area of triangle
$=\frac{1}{2}\times\text{x}\times(\text{x}\times\text{7})\text{cm}^2$
$\therefore\frac{1}{2}\times(\text{x}^2+\text{7x})=165$
$\Rightarrow x^2 + 7x - 330 = 0$
$\Rightarrow x2 + 22x - 15x - 330 = 0$
$\Rightarrow x(x + 22) - 15(x + 22) = 0$
$\Rightarrow x = -22 or x = 15 \Rightarrow x = 15$
$[\because$ base cannot be negative$]$Thus, the base of the triangle = 15m
And the altitude of triangle = (15 + 7) = 22m.
View full question & answer→Question 715 Marks
The sum of the areas of two squares is $640m^2$. If the difference in their perimeters be 64m, find the sides of the two squares.
AnswerLet x and y be the lenght of the two square fields.
$\therefore$ $x^2 + y^2 = 640 ...(1)$
4x - 4y = 64
$\therefore$ x - y = 16 ...(2)
From (2),
x = y + 16,
Putting value of x in (1)
$\Rightarrow (y + 16)^2+ y^2 = 640$
$\Rightarrow y^2 + 32y + 256 + y^2 = 640$
$\Rightarrow 2y^2 + 32y + 256 - 640 = 0$
$\Rightarrow 2y^2 + 32y - 384 = 0$
$\Rightarrow y^2 + 16y - 192 = 0$
$\Rightarrow y^2 + 24y - 8y - 192 = 0$
$\Rightarrow y(y + 24) - 8(y + 24) = 0$
$\Rightarrow (y + 24)(y - 8) = 0$
$\Rightarrow y + 24 = 0 or y - 8 = 0$
$\Rightarrow y = -24 or y = 8$
Putting y = 8 in (2)
x - 8 = 16
$\therefore$ x = 16 + 8 = 24
$\therefore$ Sides of two squares are 24m and 8m respectively.
View full question & answer→Question 725 Marks
The length of a rectangle is thrice as long as the side of a square. The side of the square is 4cm more than width of a the rectangle. Their areas being equal, find their dimensions.
AnswerLet the side of square be x cm.
Then, length of the rectangle = 3x cm
Breadth of the rectangle = (x - 4)cm
$\therefore$ Area of rectangle = Area of square x
$\therefore$ $3x(x - 4) = x^2$
$\Rightarrow 3x^2 - 12x = x^2$
$\Rightarrow 2x^2 - 12x = 0$
$\Rightarrow 2x(x - 6) = 0$
$\Rightarrow 2x = 0 or x - 6 = 0$
$\Rightarrow x = 0 or x = 6$
⇒ x = 6 (Side of the square is never 0)
Thus, side of the square = 6cm
And length of the rectangle = (3 × 6) = 18cm
Then, breadth of the rectangle = (6 - 4)cm = 2cm
View full question & answer→Question 735 Marks
A farmer prepares a rectangular vegetable garden of area 180 sq metres. With 39 metres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden.
AnswerLet the length = x meter. Area = length × breadth = 180m$^2$ Breadth $=\Big(\frac{180}{\text{x}}\Big)\text{m}$ $\therefore\text{x}=\frac{180}{\text{x}}+\frac{180}{\text{x}}=39$
$\Rightarrow x^2 - 39x + 360 = 0$
$\Rightarrow x^2 - 24x - 15x + 360 = 0$
$\Rightarrow x(x - 24) - 15(x - 24) = 0$
$\Rightarrow (x - 24)(x - 15) = 0$
$\Rightarrow x = 24 or x = 15$
If length of the rectangle = 15m Then, Breadth $=\Big(\frac{180}{15}\Big)\text{m}=12\text{m}$ Also, if length of rectangle = 24m $\therefore$ Breadth $=\frac{181}{24}\text{m}=\frac{15}2{}\text{m}=7.5\text{m}$
View full question & answer→Question 745 Marks
Solve the following equations by using the method of completing the square:
$\sqrt3\text{x}^2+10\text{x}+7\sqrt3=0$
Answer$\sqrt3\text{x}^2+10\text{x}+7\sqrt3=0$
$3\text{x}^2+10\sqrt3\text{x}+21=0$ $\big($Multiplying both sides by $\sqrt3\big)$
$\Rightarrow\text{3x}^2-10\sqrt3\text{x}=-21$
$\Rightarrow\big(\sqrt3\text{x}\big)^2-2\times\sqrt3\text{x}\times5+5^2\\=-21+5^2$ [Adding $5^2$ on both sides]
$\Rightarrow\big(\sqrt3\text{x}+5\big)^2=-21+25$
$=4=2^2$
$\Rightarrow\sqrt3\text{x}+5=\pm2$ (Taking square root on both sides)
$\Rightarrow\sqrt3\text{x}+5=2$ or $\sqrt3\text{x}+5=-2$
$\Rightarrow\sqrt3\text{x}=-3$ or $\sqrt3\text{x}=-7$
$\Rightarrow\text{x}=-\frac{3}{\sqrt3}=-\sqrt3$ or $\text{x}=-\frac{7}{\sqrt3}=-\frac{7\sqrt3}{3}$
Hence, $-\sqrt3$ and $-\frac{7\sqrt3}{3}$ are the roots of the given equation.
View full question & answer→Question 755 Marks
The sum of the squares of two consecutive positive odd numbers is 514. Find the numbers.
AnswerLet the required consecutive positive odd numbers be x and (x + 2).
Then, we have
$x^2 + (x + 2)^2 = 514$
$\Rightarrow x^2 + x^2 + 4x + 4 = 514$
$\Rightarrow 2x^2 + 4x - 510 = 0$
$\Rightarrow x^2 + 2x - 255 = 0$
$\Rightarrow x^2 - 15x + 17x - 255 = 0$
$\Rightarrow x(x - 15) + 17(x - 15) = 0$
$\Rightarrow (x - 15)(x + 17) = 0$
$\Rightarrow x - 15 = 0 or x + 17 = 0$
$\Rightarrow x = 15 or x = -17$
Since x is a positive number, x ≠ -17
⇒ x = 15
⇒ x + 2 = 15 + 2 = 147
Hence, the required positive odd numbers are 15 and 17.
View full question & answer→Question 765 Marks
Find two consecutive positive even integers whose product is 288.
AnswerLet the required consecutive positive even integers be x and (x + 2).
Then, we have
$x \times (x + 2) = 288$
$\Rightarrow x^2 + 2x - 288 = 0$
$\Rightarrow x^2 + 18x - 16x - 288 = 0$
$\Rightarrow x(x + 18) - 16(x + 18) = 0$
$\Rightarrow (x + 18)(x - 16) = 0$
$\Rightarrow x + 18 = 0 or x - 16 = 0$
$\Rightarrow x = -18 or x = 16$
Since x is a positive integer, x ≠ -18
⇒ x = 16
⇒ x + 2 = 16 + 2 = 18
Hence, the required consecutive positive even integers are 16 and 18.
View full question & answer→Question 775 Marks
The length of a hall is 3 metres more than its breadth. If the area of the hall is 238 square metres, calculate its length and breadth.
AnswerLet the breadth of hall = x meters.
Then, lenght of the hall = (x + 3) meters.
$\therefore$ Area = lenght × breadth $= 238m^2$
$\Rightarrow x \times (x + 3) = 238$
$\Rightarrow x^2 + 3x - 238 = 0$
$\Rightarrow x(x + 17) - 14(x + 17) = 0$
$\Rightarrow (x + 17)(x - 14) = 0$
$\Rightarrow x + 17 = 0 or x - 14 = 0$
$\Rightarrow x = -17 or x = 14$
⇒ x = 14 $[\because$ breadth cannot be negative$]$
Thus, the breadth of hall is 14m
And lenght of the hall is (14 + 3)m = 17m.
View full question & answer→Question 785 Marks
If the roots of the equation $(a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0$ are equal prove that $\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}.$
AnswerIt is given that the roots of the equation $(a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0$ are equal.
$\therefore$ D = 0
$\Rightarrow [-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0$
$\Rightarrow 4(a^2c^2 + b^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0$
$\Rightarrow 4(a^2c^2 + b^2d^2 + 2abcd) - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2) = 0$
$\Rightarrow (-a^2d^2 + 2abcd - b^2c^2) = 0$
$\Rightarrow -(a^2d^2 - 2abcd + b^2c^2) = 0$
$\Rightarrow (ad - bc)^2 = 0$
$\Rightarrow ad - bc = 0$
$\Rightarrow ad = bc$
$\frac{\text{a}}{\text{b}}=\frac{\text{c}}{\text{d}}.$
Hence Proved.
View full question & answer→Question 795 Marks
A rectangular field is 16m long and 10m wide. There is a path of uniform width all around it, having an area of $120m^2$. Find the width of the path.
AnswerLet the width of the path be x m Length of the field including the path = 16 + x + x = 16 + 2x
Breadth of the field including the path = 10 + x + x = 10 + 2x
Now, (Area of the field including path) - (Area of the field excluding path)= Area of the path
$\Rightarrow (16 + 2x)(10 + 2x) - (16 \times 10) = 120$
$\Rightarrow 160 + 32x + 20x + 4x^2 - 160 = 120$
$\Rightarrow 4x^2 + 52x - 120 = 0$
$\Rightarrow x^2 + 13x - 30 = 0$
$\Rightarrow x^2 + (15 - 2)x + 30 = 0$
$\Rightarrow x^2 + 15x - 2x + 30 = 0$
$\Rightarrow x(x + 15) - 2(x + 15) = 0$
$\Rightarrow (x + 15)(x - 2) = 0$
$\Rightarrow x + 15 = 0 or x - 2 = 0$
$\Rightarrow x = -15 or x = 2$
$\Rightarrow x = 2 $$[\because$ width cannot be negative$]$Thus, the width of the path is 2m.
View full question & answer→Question 805 Marks
The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2cm and exceeds twice the length of the altitude by 1cm. Find the length of each side of the triangle.
AnswerLet the base of the triangle be x Then, hypotenuse = (x + 2)cm $\therefore$ (x + 2) - (2 altitude) = 1
⇒ altitude $=\frac{1}{2}(\text{x}+1)$ By applying pythagoras theorem we have, $\therefore(\text{x}+1)^2=\text{x}^2+\frac{1}{4}(\text{x}+1)^2$ $\Rightarrow\text{x}^2+4+\text{4x}$ $=\text{x}^2+\frac{\text{x}^2}{4}+\frac{1}{4}+\frac{1}{2}\text{x}$
$\Rightarrow4+\text{4x}=\frac{\text{x}^2}{4}+\frac{1}{4}+\frac{\text{x}}2{}$+
$\Rightarrow-\frac{\text{x}^2}{4}+\frac{7}{2}\text{x}+\frac{15}{4}=0$
$\Rightarrow -x^2 + 15 + 14x = 0$
$\Rightarrow x^2 - 14x - 15 = 0$
$\Rightarrow x^2 - 15x + x - 15 = 0$
$\Rightarrow x(x - 15) + 1(x - 15) = 0$
$\Rightarrow (x - 15)(x + 1) = 0$
$\Rightarrow x = 15 or x = -1$
$\Rightarrow x = 15$
$[\because$ base cannot be negative$]$Thus, the base of the triangle be = 15m
Then, hypotenuse of triangle = (15 + 2) = 17cm
And altituted of triangle $=\frac{1}{2}(15+1)=8\text{cm}$
View full question & answer→Question 815 Marks
A two-digit number is 4 times the sum of its digits and twice the product of its digit. Find the number.
AnswerLet the tens digit be x and units digit be y.
then, 10x + y = 4(x + y) and 10x + y = 2xy
$\Rightarrow y = 2x and 10x + y = 2xy$
$Putting y = 2x in 10x + y = 2xy$
$\Rightarrow 10x + 2x = 2x.2x$
$\Rightarrow 4x^2 - 12x = 0$
$\Rightarrow 4x(x - 3) = 0$
$\Rightarrow x - 3 = 0 or x = 3$
Hence, the tens digit is 3 and units digits is (2, 3) is
Hence the required number is 36.
View full question & answer→Question 825 Marks
Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.
AnswerLet the required number be x and y.
Then,
$x^2 + y^2 = 25(x + y) ...(1)$
$x^2 + y^2 = 50(x + y) ...(2)$
$\Rightarrow 25(x + y) = 50(x - y)$
$\Rightarrow x + y = 2(x - y)$
$\Rightarrow x = 3y$
Putting x = 3y in (1), we get
$9y^2 + y^2 = 100y$
$\Rightarrow 10y^2 - 100y = 0$
$\Rightarrow 10y(y - 10) = 0$
$\Rightarrow y = 10$
Hence, x = 30 and y = 10
View full question & answer→Question 835 Marks
Solve the following quadratic equation:
$2^{2x} - 3.2^{(x+2)} + 32= 0$
Answer$2^{2x} - 3.2^{(x+2)} + 32= 0 2^{2x} - 3.2^x.2^2 + 32 = 0 y^2 - 12y + 32 = 0$
$where 2^x = y \Rightarrow y^2 - 8y - 4y + 32 = 0$
$\Rightarrow y(y - 8) - 4(y - 8) = 0$
$\Rightarrow (y - 8)(y - 4) = 0$
$\Rightarrow y - 8 = 0 or y - 4 = 0$
$\Rightarrow y = 8 or y = 4$
$\Rightarrow 2^x = 8$
$\Rightarrow 2^x = (2)^3$
$\Rightarrow x = 3$
$\Rightarrow 2^x = 4$
$\Rightarrow 2^x = (2)^2$
$\Rightarrow x = 2$
Hence, 3 and 2 are the roots of given equation.
View full question & answer→Question 845 Marks
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes. To reach the destination, 1500km away, in time, the pilot increased the speed by 100km/hour.
Find the original speed of the plane.
Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?
AnswerLet the original speed of the train be x km/hr Then, Time taken to cover 1500km with the original speed $=\frac{1500}{\text{x}}\text{hrs}$ Time taken to cover 1500km with the speed of (x + 100)km/hr
$=\frac{1500}{\text{x}+100}\text{hrs}$
$\therefore\frac{1500}{\text{x}}=\frac{1500}{\text{x}+100}+\frac{1}{2}$
$\Rightarrow\frac{1500}{\text{x}}-\frac{1500}{\text{x}+100}=\frac{1}{2}$
$\Rightarrow\frac{1500\text{x}+150000-1500\text{x}}{\text{x}^2+\text{100x}}=\frac{1}{2}$
$\Rightarrow 300000 = x^2 + 100x$
$\Rightarrow x^2 + 100x - 300000 = 0$
$\Rightarrow x^2 + 600x - 500x - 300000 = 0$
$\Rightarrow x(x + 600) - 500(x + 600) = 0$
$\Rightarrow (x + 600)(x - 500) = 0$
$\Rightarrow x + 600 = 0 or x - 500 = 0$
$\Rightarrow x = -600 or x = 500$
Since speed cannot be negative,
$\text{x}\neq-600.$
⇒ x = 500 Hence, the original speed of the plane is 500km/hr.Yes, we apperciate the value shown by the pilot.
View full question & answer→Question 855 Marks
A train covers a distance of 300km at a uniform speed. If the speed of the train is increased by 5 km/hr, it take 2 hours less in the journey. Find the original speed of the train.
AnswerDistance travelled = 300km
Let the original speed of train be x kmph
Recall that, time taken $=\frac{\text{Distance}}{\text{Speed}}$
Hence time taken $=\frac{300}{\text{x}}$
Given that the speed of the train is increased by 5km an hour.
Hence the new speed is (x + 5)kmph
Time taken to cover 300km $=\frac{300}{(\text{x}+5)}$
Given that the time taken 2 hours less when compared to the previous time.
$\Rightarrow\frac{300}{\text{x}}-\frac{300}{\text{x}+5}=2$
$\Rightarrow\frac{300(\text{x}+5)-300\text{x}}{\text{x}(\text{x+5})}=2$
$\Rightarrow 300x + 1500 - 300x = 2x^2 + 10x$
$\Rightarrow 2x^2 + 10x - 1500 = 0$
$\Rightarrow x^2 + 5x - 750 = 0$
$\Rightarrow x^2 + 30x - 25x - 750 = 0$
$\Rightarrow x(x + 30) - 25(x + 30) = 0$
$\Rightarrow (x + 30)(x - 25) = 0$
$\Rightarrow x + 30 = 0 or x - 25 = 0$
$\therefore$ $x = -30 or x = 25$
Since speed cannot be negative, hence the original speed of train is 25kmph.
View full question & answer→Question 865 Marks
The sum of a natural number and its reciprocal is $\frac{65}{8}.$ Find the number.
AnswerLet the required numbers be x.
Then, its reciprocal $=\frac{1}{\text{x}}$
Thus, we have
$\text{x}+\frac{1}{\text{x}}=\frac{65}{8}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=\frac{65}8{}$
$\Rightarrow 8x^2+ 8 = 65x$
$\Rightarrow 8x^2 - 65x + 8 = 0$
$\Rightarrow 8x^2 - 64x - x + 8 = 0$
$\Rightarrow 8x(x - 8) - 1(x - 8) = 0$
$\Rightarrow (x - 8)(8x - 1) = 0$
$\Rightarrow x - 8 = 0 or 8x - 1 = 0$
⇒ x = 8 or $\text{x}=\frac{1}{8}$
Since x is a natural number, $\text{x}\neq\frac{1}{8}.$
⇒ x = 8
Hence, the required number is 8.
View full question & answer→Question 875 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$
AnswerGiven: $\sqrt3\text{x}^2-2\sqrt2\text{x}-2\sqrt3=0$ On comparing it with $ax^2 + bx + c = 0,$ we get:
$\text{x}=\sqrt3,\ \text{b}=-2\sqrt2$ and $\text{c}=-2\sqrt3$
$\therefore$ Discriminant D is given by: $D = (b^2 - 4ac)$
$=\big(-2\sqrt2\big)^2-4\times\sqrt3\times\big(-2\sqrt3\big)$
$=8+24$ $=32>0$ So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{32}=4\sqrt2$ $\therefore\ \alpha=\frac{-\text{b}+\sqrt{\text{D}}}{\text{2a}}$ $=\frac{-\big(-2\sqrt2\big)+4\sqrt{2}}{2\times\sqrt3}$
$=\frac{6\sqrt2}{2\sqrt3}$
$=\sqrt6$ $\beta=\frac{-\text{b}-\sqrt{\text{D}}}{\text{2a}}$
$=\frac{-\big(-2\sqrt2\big)-4\sqrt{2}}{2\times\sqrt3}$
$=\frac{-2\sqrt2}{2\sqrt3}$
$=-\frac{\sqrt6}{3}$
Hence, $\sqrt6$ and $-\frac{\sqrt6}{3}$ are the roots of the given equation.
View full question & answer→Question 885 Marks
Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
AnswerLet the smaller part and larger part be x, 16 - x.
Then,
$\Rightarrow 2x^2- (16 - x)^2 = 164$
$\Rightarrow 2x^2 - (256 + x^2 - 32x) = 164$
$\Rightarrow 2x^2 - 256 - x^2 + 32x = 164$
$\Rightarrow x^2 + 32x - 256 - 164 = 0$
$\Rightarrow x^2 + 32x - 420 = 0$
$\Rightarrow x^2 + 42x - 10x - 420 = 0$
$\Rightarrow x(x + 42) - 10(x + 42)$
$\Rightarrow x + 42 = 0 or x - 10 = 0$
$\Rightarrow x = -42 or x = 10$
-42 is not a positive part
Hence, the larger and smaller parts are 10, 6 respectively.
View full question & answer→Question 895 Marks
The length of a rectangular field is three times its breadth. If the area of the field by 147 sq metres, find the length of the field.
AnswerLet the breadth of a rectangle = x meter
Then, lenght of the rectangle = 3x meter
$\therefore$ Area = lenght × breadth $= 147cm^2$
$\Rightarrow x \times 3x = 147$
$\Rightarrow 3x^2 = 147$
$\Rightarrow x^2 = 49$
$\Rightarrow\text{x}=\sqrt{49}$
$\Rightarrow\text{x}=\pm7$
$\Rightarrow\text{x}=7$ or $ \text{x}=-7$
$\Rightarrow\text{x}=7$ $[\because$ breadth cannot be negative$]$
Thus, breadth of rectangle = 7m
And lenght of rectangle = (3 × 7)m = 21m.
View full question & answer→Question 905 Marks
The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is $2\frac{9}{10}.$ Find the fraction.
AnswerLet the numerator and denominator be x. x+ 3
Then,
$\frac{\text{x}}{(\text{x}+3)}+\frac{(\text{x}+3)}{\text{x}}=2\frac{9}{10}$
$\Rightarrow\frac{\text{x}^2+(\text{x}+3)^2}{\text{x}(\text{x}+3)}=\frac{29}{10}$
$\Rightarrow\frac{\text{x}^2+\text{x}^2+9+\text{6x}}{\text{x}^2+\text{3x}}=\frac{{29}}{10}$
$\Rightarrow 20x^2 + 90 + 60x = 29x^2 + 87x$
$\Rightarrow 20x^2 - 29x^2 + 60x - 87x + 90 = 0$
$\Rightarrow -9x^2- 27x + 90 = 0$
$\Rightarrow x^2 + 3x - 10 = 0$
$\Rightarrow x^2 + (5x - 2x) - 10 = 0$
$\Rightarrow x(x + 5) - 2(x + 5) = 0$
$\Rightarrow (x + 5)(x - 2) = 0$
$\Rightarrow x + 5 = 0 or x - 2 = 0$
$\Rightarrow x = -5 or x = 2$
Hence, numerator and denominator are 2 and 5 respectively and fraction is $\frac{2}{5}.$
View full question & answer→Question 915 Marks
The sum of a number and its reciprocal is $2\frac{1}{30}.$ Find the number.
AnswerLet the required number be x.
Then, its reciprocal $=\frac{1}{\text{x}}$
Thus, we have
$\text{x}+\frac{1}{\text{x}}=2\frac{1}{30}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=\frac{61}{30}$
$\Rightarrow 30x^2 + 30 = 61x$
$\Rightarrow 30x^2 - 61x + 30 = 0$
$\Rightarrow 30x^2- 36x - 25x + 30 = 0$
$\Rightarrow 6x(5x - 6) - 5(5x - 6) = 0$
$\Rightarrow (5x - 6)(6x - 5) = 0$
$\Rightarrow 5x - 6 = 0 or 6x - 5 = 0$
$\Rightarrow\text{x}=\frac{6}{5}$ or $\text{x}=\frac{5}{6}$
Hence, the required number is $\frac{6}{5}$ or $\frac{5}{6}.$
View full question & answer→Question 925 Marks
A train travels at a certain average speed for a distance of 54km and then travels a distance of 63km at an average of 6km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?
AnswerLet x be the first speed of the train.
We know that $\frac{\text{Distance}}{\text{Speed}}=\text{Time}$
Thus, we have
$\frac{54}{\text{x}}+\frac{63}{\text{x}+6}=3\ \text{hours}.$
$\Rightarrow\frac{54(\text{x}+6)+63\text{x}}{\text{x}(\text{x}+6)}=3$
$\Rightarrow 54(x + 6) + 63x = 3x(x + 6)$
$\Rightarrow 54x + 324 + 63x = 3x^2 + 18x$
$\Rightarrow 117x + 324 = 3x^2 + 18x$
$\Rightarrow 3x^2 - 117x - 324 + 18x = 0$
$\Rightarrow 3x^2 - 99x - 324 = 0$
$\Rightarrow x^2 - 33x - 108 = 0$
$\Rightarrow x^2 - 36x + 3x - 108 = 0$
$\Rightarrow x(x - 36) + 3(x - 36) = 0$
$\Rightarrow (x - 36)(x + 3) = 0$
$\Rightarrow x - 36 = 0 or x + 3 = 0$
$\Rightarrow x = 36 or x = -3$
Since speed cannot be negative
Hence, the intial speed of the train is 36km/hr.
View full question & answer→Question 935 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$5x^2 - kx + 1 = 0$
AnswerThe given equation is $5x^2 - kx + 1 = 0$
$\therefore$$ D = (-k)^2 - 4 \times 5 \times 1$
$D = k^2 - 20$
The given equation has real and distinct roots if D > 0.
$\therefore\text{k}^2-20>0$
$\Rightarrow\text{k}^2-\big(2\sqrt5\big)^2>0$
$\Rightarrow\big(\text{k}-2\sqrt5\big)\big(\text{k}+2\sqrt5\big)>0$
$\Rightarrow\text{k}<-2\sqrt5$ or $\text{k}>2\sqrt5$
View full question & answer→Question 945 Marks
A motorboat whose speed is 9km/hr in still water, goes 15km downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.
AnswerLet the speed of the stream be = x km/h.Speed of the boat in still water = 9km/h
Speed of the boat down stream = 9 + x km/h
$\therefore$ time taken by boat to go 15km downstream $=\frac{15}{9+\text{x}}\text{hrs.}$
Speed of boat upstream = 9 - x
$\therefore$ time taken by boat to go 15km up stream $=\frac{15}{9-\text{x}}\text{hrs.}$
Total time $=\frac{15}{9+\text{x}}+\frac{15}{9-\text{x}}=3\frac{45}{60}=3\frac{3}{4}=\frac{15}{4}$
Dividing by 15
$\frac{1}{9+\text{x}}+\frac{1}{9-\text{x}}=\frac{1}{4}$
$\Rightarrow\frac{9+\text{x}+9-\text{x}}{(9+\text{x})(9-\text{x})}=\frac{1}{4}$
$\Rightarrow\frac{18}{81-\text{x}^2}=\frac{1}{4}$
$\Rightarrow81-\text{x}^2=72$
$\therefore\text{x}^2=81-72=9$
$\therefore\text{x}=\pm3$
But $\text{x}\neq-3$
$\therefore$ Speed of the stream = 3km/h.
View full question & answer→Question 955 Marks
The perimeter of a rectangular plot is 62m and its area is 228 sq metres. Find the dimensions of the plot.
AnswerPerimeter of a rectangle = 62m.
Let the lenght of the rectangle be x m. Then, breadth of the rectangle
$=\frac{\text{Perimeter}}{2}-\text{Length}$
$=\frac{62}{2}-\text{x}=(31-\text{x})\text{m}$Now, Area $= 228m^2$
Area = lenght × breadth $= 228m^2$
$\Rightarrow x(31 - x) = 228$
$\Rightarrow 31x - x^2 = 228$
$\Rightarrow x^2 - 31x + 228 = 0$
$\Rightarrow x^2 + 3x - 238 = 0$
$\Rightarrow x^2 - 19x - 12x + 228 = 0$
$\Rightarrow x(x - 19) - 12(x - 19) = 0$
$\Rightarrow (x - 19)(x - 12) = 0$
$\Rightarrow x - 19 = 0 or x - 12 = 0$
$\Rightarrow x = 19 or x = -12$
$\Rightarrow x = 19$
$[\because$ breadth cannot be negative$]$
Hence, the lenght of a rectangle is 19m and the breadth of a rectangle is 12m.
View full question & answer→Question 965 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\text{x}^2-(\sqrt3+1)\text{x}+\sqrt3=0$
AnswerThe given equation is:$\text{x}^2-(\sqrt3+1)\text{x}+\sqrt3=0$
Comparing it with $ax^2 + bx + c = 0$, we get
$\text{a}=1,\ \text{b}=-(\sqrt3+1)$ and $\text{c}=\sqrt3$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=[-(\sqrt3+1)]^2-4\times1\times\sqrt3$
$=3+1+2\sqrt3-4\sqrt3$
$=3-2\sqrt3+1$
$=\big(\sqrt3-1\big)^2>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{\big(\sqrt3-1\big)^2}=\sqrt3-1$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]+\big(\sqrt3-1\big)}{2\times1}$
$=\frac{\sqrt3+1+\sqrt3-1}{2}$
$=\frac{2\sqrt3}{2}$
$=\sqrt3$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-\big[-\big(\sqrt3+1\big)\big]-\big(\sqrt3-1\big)}{2\times1}$
$=\frac{\sqrt3+1-\sqrt3+1}{2}$
$=\frac{2}{2}$
$=1$
Hence, $\sqrt3$ and 1 are the roots of the given equation.
View full question & answer→Question 975 Marks
If a and b are real and a ≠ b then show that the roots of the equation $(a - b)x^2 + 5(a + b)x - 2(a - b) = 0$ are real and unequal.
AnswerThe given equation is $(a-b) x^2+5(a+b) x-2(a-b)=0$
$\therefore D=[5(a+b)]^2-4 \times(a-b) \times[-2(a-b)]$
$D=25(a+b)^2+8(a-b)^2$
Since $a$ and $b$ are real and $a \neq b$, so $(a-b)^2>0$ and $(a+b)^2>0$.
$\therefore 8(a-b)^2>0 \ldots(1)$ (Product of two positive numbers is always positive)
Also, $25(a+b)^2>0 \ldots(2)$ (Product of two positive numbers is always positive)
Adding (1) and (2), we get
$25(a+b)^2+8(a-b)^2>0$ (Sum of two positive numbers is always positive)
$\Rightarrow D>0$
Hence, the roots of the given equation are real and unequal.
View full question & answer→Question 985 Marks
One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. Find their present ages.
AnswerLet the age of the son be x and age of man be y. 1 year ago
$\therefore$ (y - 1) = 8(x - 1)
$\Rightarrow y - 1 = 8x - 8$
$\Rightarrow y - 8x = -7$
$\Rightarrow y = -7 + 8x Also, y = x^2$
$\Rightarrow (-7 + 8x) = x^{2$
$} \Rightarrow 8x - 7 = x^2$
$\Rightarrow x^2 - 8x + 7 = 0$
$\Rightarrow x^2 - 7x - x + 7 = 0$
$\Rightarrow x(x - 7) - 1(x - 7) = 0$
$\Rightarrow (x - 7)(x - 1) = 0$
$\Rightarrow x - 7 = 0 or x - 1 = 0$
$\Rightarrow x = 7 or x = 1$
⇒ x = 7 $[\text{x}\neq1]$ Age of the son = 7 years.
Age of man = (-7 + 8 × 7) = -7 + 56 = 49 years.
View full question & answer→Question 995 Marks
The sum of two natural numbers is 9 and the sum of their reciprocals is $\frac{1}{2}.$ Find the numbers.
AnswerLet the required numbers be x and (9 - x).
Then, we have
$\frac{1}{\text{x}}+\frac{1}{9-\text{x}}=\frac{1}{2}$
$\Rightarrow\frac{9-\text{x}+\text{x}}{\text{x}(9-\text{x})}=\frac{1}{2}$
$\Rightarrow 18 = 9x - x^2$
$\Rightarrow x^2 - 9x + 18 = 0$
$\Rightarrow x^2 - 6x - 3x + 18 = 0$
$\Rightarrow x(x - 6) - 3(x - 6) = 0$
$\Rightarrow (x - 6)(x - 3) = 0$
$\Rightarrow x - 6 = 0 or x - 3 = 0$
$\Rightarrow x = 6 or x = 3$
Hence, the required numbers are 3 and 6.
View full question & answer→Question 1005 Marks
Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
AnswerLet the time taken by the smaller pipe to fill the tank be x hours.
Then, the time taken by the larger pipe = (x - 9) hours.
Part of tank filled by smaller pipe in 1 hour $=\frac{1}{\text{x}}$
Part of tank filled by smaller pipe in 1 hour $=\frac{1}{\text{x}-9}$
It is given that the tank can be filled in 6 hours by both the pipes together.
$\therefore\frac{1}{\text{x}}+\frac{1}{\text{x}-9}=\frac{1}{6}$
$\Rightarrow\frac{100}{\text{x}}+\frac{100}{\text{x}+5}=9$
$\Rightarrow\frac{\text{x}-9+\text{x}}{\text{x}^2-\text{9x}}=\frac{1}{6}$
$\Rightarrow 6(2x - 9) = x^2 - 9x$
$\Rightarrow 12x - 54 = x^2- 9x$
$\Rightarrow x^2 - 21x + 54 = 0$
$\Rightarrow x^2 - 18x + 3x + 54 = 0$
$\Rightarrow x(x - 18) - 3(x - 18) = 0$
$\Rightarrow (x - 18)(x - 3) = 0$
$\Rightarrow x - 18 = 0 or x - 3 = 0$
$\Rightarrow x = 18 or x = 3$
Since time cannot be less than 9, $\text{x}\neq3$
⇒ x = 18
⇒ x - 9 = 18 - 9 = 9
Hence, the smaller pipe taken 18 hours to fills the tank and the larger pipe takes 9 hours to fills the tank.
View full question & answer→Question 1015 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\text{x}-\frac{1}{\text{x}}=3,\ \text{x}\neq0$
AnswerThe given equation is$\text{x}-\frac{1}{\text{x}}=3,\ \text{x}\neq0$
$\Rightarrow\frac{\text{x}^2-1}{\text{x}}=3$
$\Rightarrow\text{x}^2-1=3\text{x}$
$\Rightarrow\text{x}^2-3\text{x}-1=0$
This equation is of the form $ax ^2+ bx + c =0$,
where $a =1, b=-3$ and $c =-1 . $
$\therefore$ Discriminant,
$D = b ^2-4 ac$ $=(-3)^2-4 \times 1 \times(-1)=9+4=13>0$
So, the given equation has real roots. Now,
$\sqrt{\text{D}}=\sqrt{13}$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-3)+\sqrt{13}}{2\times1}$
$=\frac{3+\sqrt{13}}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-3)-\sqrt{13}}{2\times1}$
$=\frac{3-\sqrt{13}}{2}$
Hence, $\frac{3+\sqrt{13}}{2}$ and $\frac{3-\sqrt{13}}{2}$ are the roots of the given equation.
View full question & answer→Question 1025 Marks
Solve the following quadratic equation:
$\frac{4}{\text{x}}-3=\frac{5}{\text{2x}+3},$ $\text{x}\neq=0,\ \frac{-3}{2}$
Answer$\frac{4}{\text{x}}-3=\frac{5}{\text{2x}+3}$ $\Rightarrow\frac{4-\text{3x}}{\text{x}}=\frac{5}{\text{2x}+3}$⇒ (4 - 3x)(2x + 3) = 5x
$\Rightarrow 8x + 12 - 6x^2 - 9x = 5x$
$\Rightarrow 12 - 6x^2 - x = 5x$
$\Rightarrow 12 - 6x^2 - 6x = 0$
$\Rightarrow 6x^2 + 6x - 12 = 0$
$\Rightarrow x^2 + x - 2 = 0$
$\Rightarrow x^2 + 2x - x - 2 = 0$
$\Rightarrow x(x + 2) - 1(x + 2) = 0$
$\Rightarrow (x + 2)(x - 1) = 0$
$\Rightarrow x + 2 = 0 or x - 1 = 0$
$\Rightarrow x = -2 or x = 1$
View full question & answer→Question 1035 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:$12abx^2 - (9a^2 - 8b^2)x - 6ab = 0,$ where $ \text{a}\neq0$ and $\text{b}\neq0$
AnswerGiven, $12 a b x^2-\left(9 a^2-8 b^2\right) x-6 a b=0$
On comparing it with $A x^2+B x+C=0$, we get:
$A=12 a b, B=-\left(9 b^2-8 b^2\right)$ and $C=-6 a b$
Discriminant $D$ is given by:
$D=B^2-4 A C$
$=\left[-\left(9 a^2-8 b^2\right)\right]^2-4 \times 12 a b \times(-6 a b)$
$=81 a^4-144 a^2 b^2+64 b^4+288 a^2 b^2$
$=81 a^4+144 a^2 b^2+64 b^4$
$=\left(9 a^2+8 b^2\right)>0$
Hence, the roots of the equation are equal.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(9\text{a}^2-8\text{b}^2)]+\sqrt{(9\text{a}^2+8\text{b}^2)^2}}{2\times12\text{a}\text{b}}$
$=\frac{9\text{a}^2-8\text{b}^2+9\text{a}^2+8\text{b}^2}{24\text{ab}}$
$=\frac{18\text{a}^2}{24\text{ab}}$
$=\frac{3\text{a}}{4\text{b}}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(9\text{a}^2-8\text{b}^2)]-\sqrt{(9\text{a}^2+8\text{b}^2)^2}}{2\times12\text{ab}}$
$=\frac{9\text{a}^2-8\text{b}^2-9\text{a}^2-8\text{b}^2}{24\text{ab}}$
$=\frac{-16\text{b}^2}{24\text{ab}}$
$=\frac{-2\text{b}}{3\text{a}}$
Thus, the roots of the equation are $\frac{3\text{a}}{4\text{b}}$ and $\frac{-2\text{b}}{3\text{a}}.$
View full question & answer→Question 1045 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula: $a^2 b^2 x^2-\left(4 b^4-3 a^4\right) x-12 a^2 b^2=0, a \neq 0$ and $b \neq 0$
AnswerThe Given equation is $a^2 b^2 x^2-\left(4 b^4-3 a^4\right) x-12 a^2 b^2=0$. Comparing it with $A x^2+B x+C=0$, we get $A=a^2 b^2, B=-\left(4 b^4-3 a\right)$ and $C=-12 a 2 b^2$
$\therefore$ Discriminant, $D = B ^2-4 AC$
$= [-(4b^4 - 3a^4)]^2 - 4 \times a^2b^2 \times (-12a^2b^2)$
$= 16b^8 - 24a^4b^4 + 9a^8 + 48a^4b^4$
$=16b^8 + 24a^4b^4 + 9a^8$
$= (4b^4 + 3a^4) > 0$
So, the given equation has real rooots.
Now, $\sqrt{\text{D}}=\sqrt{-(4\text{b}^4-3\text{a}^4)^2}$
$=4\text{b}^4+3\text{a}^4$
$\therefore$ $\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(4\text{b}^4-3\text{a}^4)]+(4\text{b}^4+3\text{a}^4)}{2\times\text{a}^2\text{b}^2}$
$=\frac{8\text{b}^4}{2\text{a}^2\text{b}^2}$
$=\frac{4\text{b}^2}{\text{a}^2}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(4\text{b}^4-3\text{a}^4)]-(4\text{b}^4+3\text{a}^4)}{2\times\text{a}^2\text{b}^2}$
$=\frac{-6\text{a}^4}{2\text{a}^2\text{b}^2}$
$=\frac{-3\text{a}^2}{\text{b}^2}$
Hence, $\frac{4\text{b}^2}{\text{a}^2}$ and $-\frac{3\text{a}^2}{\text{b}^2}$ are the roots of the given equation.
View full question & answer→Question 1055 Marks
Solve the following quadratic equation:
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3},$ $\text{x}\neq2,4$
Answer$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3}$
$\Rightarrow\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=\frac{10}{3}$
$\Rightarrow\frac{(\text{x}-1)(\text{x}-4)+(\text{x}-3)(\text{x}-2)}{(\text{x}-2)(\text{x}-4)}=\frac{10}{3}$
$\Rightarrow\frac{(\text{x}^2-\text{4x}-\text{x}+4)+(\text{x}^2-\text{2x}-\text{3x}+6)}{\text{x}^2-\text{4x}-\text{2x}+8}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-5\text{x}+4+\text{x}^2-5\text{x}+6}{\text{x}^2-6\text{x}+8}=\frac{10}{3}$
$\Rightarrow\frac{\text{2x}^2-\text{10x}+10}{\text{x}^2-6\text{x}+8}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-\text{5x}+5}{\text{x}^2-6\text{x}+8}=\frac{5}{3}$
$\Rightarrow 3x^2 - 15x + 15 = 5x^2 - 30x + 40$
$\Rightarrow 2x^2 - 15x + 25 = 0$
$\Rightarrow 2x^2 - 10x - 5x + 25 = 0$
$\Rightarrow 2x(x - 5) - 5(x - 5) = 0$
$\Rightarrow (x - 5)(2x - 5) = 0$
$\Rightarrow x - 5 = 0 or 2x - 15 = 0$
⇒ x = 5 or $\text{x}=\frac{5}{2}$
View full question & answer→Question 1065 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$4x^2 - 4a^2x + (a^4 - b^4) = 0$
AnswerThe given equation is $4 x^2-4 a^2 x+\left(a^4-b^4\right)=0$.
Comparing it with $Ax ^2+ Bx + C =0$, we get
$A=4, B=-4 a^2$ and $C=a^4-b^4$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= (-4a^2)^2 - 4 \times 4 \times (a^4 - b^4)$
$= 16a^4 - 16a^4 + 16b^4$
$= 16b^4 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{16\text{b}^4}=4\text{b}^2$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-4\text{a}^2)+4\text{b}^2}{2\times4}$
$=\frac{4(\text{a}^2+\text{b}^2)}{8}$
$=\frac{\text{a}^2+\text{b}^2}{2}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-4\text{a}^2)-4\text{b}^2}{2\times4}$
$=\frac{4(\text{a}^2-\text{b}^2)}{8}$
$=\frac{\text{a}^2-\text{b}^2}{2}$
Hence, $\frac{1}{2}(\text{a}^2+\text{b}^2)$ and $\frac{1}{2}(\text{a}^2-\text{b}^2)$ are the roots of the given equation.
View full question & answer→Question 1075 Marks
Solve the following quadratic equation:
$\Big(\frac{\text{4x}-3}{\text{2x}+1}\Big)-10\Big(\frac{\text{2x}+1}{\text{4x}-3}\Big)=3,$ $\text{x}\neq\frac{-1}{2},\ \frac{3}{4}$
Answer$\Big(\frac{\text{4x}-3}{\text{2x}+1}\Big)-10\Big(\frac{\text{2x}+1}{\text{4x}-3}\Big)=3$ Taking $\frac{\text{4x}-3}{\text{2x}+1}=\text{y},$
we have $\text{y}-\frac{10}{\text{y}}=3$
$\Rightarrow\frac{\text{y}^2-10}{\text{y}}=3$
$\Rightarrow y^2 - 10 = 3y$
$\Rightarrow y^2 - 3y - 10 = 0$
$\Rightarrow y^2 - 5y + 2y - 10 = 0$
$\Rightarrow y(y - 5) + 2(y - 5) = 0$
$\Rightarrow (y - 5)(y + 2) = 0$
$\Rightarrow y - 5 = 0 or y + 2 = 0$
$\Rightarrow y = 5 or y = -2$
$\Rightarrow\frac{\text{4x}-3}{\text{2x}+1}=5$ or $\frac{\text{4x}-3}{\text{2x}+1}=-2$
$\Rightarrow 4x - 3 = 10x + 5 or 4x - 3 = -4x - 2$
$\Rightarrow 6x = -8 or 8x = 1$
$\Rightarrow\text{x}=\frac{-4}{3}$ or $\text{x}=\frac{1}{8}$
View full question & answer→Question 1085 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$x^2 - 4ax - b^2 + 4a^2 = 0$
AnswerThe given equation is $x^2 - 4ax - b^2 + 4a^2 = 0$
. Comparing it with $Ax^2 + Bx + C = 0$,
we get A = 1, B = -4a and $C = -b^2+ 4a^2$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= (-4a)^2 - 4 \times 1 \times (-b^2 + 4a^2) = 16a^2 + 4b^2 - 16a^2 = 4b^2 > 0 $So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{4\text{b}^2}=2\text{b}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-4\text{a})+2\text{b}}{2\times1}$
$=\frac{4\text{a}+2\text{b}}{2}$
$=2\text{a}+\text{b}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-4\text{a})-2\text{b}}{2\times1}$
$=\frac{4\text{a}-2\text{b}}{2}$
$=2\text{a}-\text{b}$
Hence, (2a + b) and (2a - b) are the roots of the given equation.
View full question & answer→Question 1095 Marks
Solve the following quadratic equation:
$\frac{\text{x}-4}{\text{x}-5}+\frac{\text{x}-6}{\text{x}-7}=3\frac{1}{3},$ $\text{x}\neq5,7$
Answer$\frac{\text{x}-4}{\text{x}-5}+\frac{\text{x}-6}{\text{x}-7}=3\frac{1}{3}$
$\Rightarrow\frac{\text{x}-4}{\text{x}-5}+\frac{\text{x}-6}{\text{x}-7}=\frac{10}{3}$
$\Rightarrow\frac{(\text{x}-4)(\text{x}-7)+(\text{x}-6)(\text{x}-5)}{(\text{x}-5)(\text{x}-7)}=\frac{10}{3}$
$\Rightarrow\frac{(\text{x}^2-\text{7x}-\text{4x}+28)+(\text{x}^2-\text{5x}-\text{6x}+30)}{\text{x}^2-\text{7x}-\text{5x}+35}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-11\text{x}+28+\text{x}^2-11\text{x}+30}{\text{x}^2-12\text{x}+35}=\frac{10}{3}$
$\Rightarrow\frac{\text{2x}^2-\text{22x}+58}{\text{x}^2-12\text{x}+35}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-\text{11x}+29}{\text{x}^2-12\text{x}+35}=\frac{5}{3}$
$\Rightarrow 3x^2 - 33x + 87 = 5x^2 - 60x + 175$
$\Rightarrow 2x^2 - 27x + 88 = 0$
$\Rightarrow 2x^2 - 16x - 11x + 88 = 0$
$\Rightarrow 2x(x - 8) - 11(x - 8) = 0$
$\Rightarrow (x - 8)(2x - 11) = 0$
$\Rightarrow x - 8 = 0 or 2x - 11 = 0$
⇒ x = 8 or $\text{x}=\frac{11}{2}$
⇒ x = 8 or $\text{x}=5\frac{1}{2}$
View full question & answer→Question 1105 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 + 6x - (a^2 + 2a -8) = 0$
AnswerThe given equation is $x^2+6 x-\left(a^2+2 a-8\right)=0$
Comparing it with $A x^2+B x+C=0$, we getA $=1, B=6$ and $C=-\left( a ^2+2 a -8\right)$
\$itherefore\$ Discriminant, $D=B^2-4 A C$
$=6^2-4 \times 1 \times\left[-\left(a^2+2 a-8\right)\right]=36+4 a^2+8 a-32=4 a^2+8 a+4=4\left(a^2+2 a+1\right)=4(a+1)^2>0$ So, the given equation has real
roots.Now, $\sqrt{\text{D}}=\sqrt{4(\text{a}+1)^2}=2(\text{a}+1)$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-6+2(\text{a}+1)}{2\times1}$
$=\frac{2\text{a}-4}{2}$
$=\text{a}-2$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-6-2(\text{a}+1)}{2\times1}$
$=\frac{-2\text{a}-8}{2}$
$=-\text{a}-4$
$=-(\text{a}+4)$
Hence, (a - 2) and -(a + 4) are the roots of the given equation.
View full question & answer→Question 1115 Marks
Solve the following quadratic equation:
$9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0$
Answer$9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2)$
$\Rightarrow 9x^2 - 9(a + b)x + (2a^2 + 4ab + ab + 2b^2) = 0$
$\Rightarrow 9x^2 - 9(a + b)x + [2a(a + 2b) + b(a + 2b)] = 0$
$\Rightarrow 9x^2 - 9(a + b)x + (a + 2b)(2a + b) = 0$
$\Rightarrow 9x^2 - 3(a + 2b)x - 3(2a + b)x + (a + 2b)(2a + b) = 0$
$\Rightarrow 3x[3x - (a + 2b)] - (2a + b)[3x - (a + 2b)] = 0$
$\Rightarrow [3x - (a + 2b)][3x - (2a + b)] = 0$
$\Rightarrow 3x - (a + 2b) = 0 or 3x - (2a + b) = 0$
$\Rightarrow\text{x}=\frac{\text{a}+\text{2b}}{3}$ or $\text{x}=\frac{\text{2a}+\text{b}}{3}$
View full question & answer→Question 1125 Marks
Solve the following quadratic equation:
$\frac{1}{\text{2x}-3}+\frac{1}{\text{x}-5}=1\frac{1}{9},$ $\text{x}\neq\frac{3}{2},\ 5$
Answer$\frac{1}{\text{2x}-3}+\frac{1}{\text{x}-5}=1\frac{1}{9}$
$\Rightarrow\frac{1}{\text{2x}-3}+\frac{1}{\text{x}-5}=\frac{10}{9}$
$\Rightarrow\frac{\text{x}-5+\text{2x}-3}{(\text{2x}-3)(\text{x}-5)}=\frac{10}{9}$
$\Rightarrow\frac{\text{3x}-8}{(\text{2x}-3)(\text{x}-5)}=\frac{10}{9}$
$\Rightarrow 27x - 72 = 10[(2x - 3)(x - 5)]$
$\Rightarrow 27x - 72 = 10[2x^2 - 10x - 3x + 15]$
$\Rightarrow 27x - 72 = 10[2x^2 - 13x + 15]$
$\Rightarrow 27x - 72 = 20x^2 - 130x + 150$
$\Rightarrow 20x^2 - 157x + 222 = 0$
$So, a = 20, b = -157, c = 222$
So, $\text{x}=\frac{157\pm\sqrt{(-157)^2-4(20)(222)}}{40}$
$\Rightarrow\text{x}=\frac{157\pm\sqrt{24649-17760}}{40}$
$\Rightarrow\text{x}=\frac{157\pm\sqrt{6889}}{40}$
$\Rightarrow\text{x}=\frac{157\pm83}{40}$
$\Rightarrow\text{x}=\frac{157+83}{40}$ or $\text{x}=\frac{157-83}{40}$
$\Rightarrow\text{x}=6$ or $\text{x}=1.85$
View full question & answer→Question 1135 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$4x^2 + 4bx - (a^2 - b^2) = 0$
AnswerThe given equation is $4 x^2+4 b x-\left(a^2-b^2\right)=0$. Comparing it with $Ax ^2+ Bx + C =0$, we get $A=4, B=4 b$ and $C=-\left(a^2-b^2\right)$
$\therefore$ Discriminant, $D = B ^2-4 AC$
$= (4b)^2 - 4 \times 4 \times [-(a^2 - b^2)]$
$= 16b^2 + 16a^2 -16b^2$
$= 16a^2 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{16\text{a}^2}=4\text{a}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-4\text{b}+4\text{a}}{2\times4}$
$=\frac{4(\text{a}-\text{b})}{8}$
$=\frac{\text{a}-\text{b}}{2}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-4\text{b}-4\text{a}}{2\times4}$
$=\frac{-4(\text{a}+\text{b})}{8}$
$=-\frac{\text{a}+\text{b}}{2}$
Hence, $\frac{1}{2}(\text{a}-\text{b})$ and $-\frac{1}{2}(\text{a}+\text{b})$ are the roots of the given equation.
View full question & answer→Question 1145 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$\text{x}+\frac{1}{\text{x}}=3,\ \text{x}\neq0$
AnswerThe given equation is:$\text{x}+\frac{1}{\text{x}}=3,\ \text{x}\neq0$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=3$
$\Rightarrow\text{x}^2+1=3\text{x}$
$\Rightarrow\text{x}^2-3\text{x}+1=0$
The equation is of the form $ax^2 + bx + c = 0,$ where a = 1, b = -3 and c = 1.$\therefore$ Discriminant, $D = b^2- 4ac=(-3)^2 - 4 \times 1 \times 1$
= 9 - 4 = 5 > 0
So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt5$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-3)+\sqrt5}{2\times1}$
$=\frac{3+\sqrt5}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-(-3)-\sqrt5}{2\times1}$
$=\frac{3-\sqrt5}{2}$
Hence, $\frac{3+\sqrt5}{2}$ and $\frac{3-\sqrt5}{2}$ are the roots of the given equation.
View full question & answer→Question 1155 Marks
Solve the following quadratic equation:
$\frac{\text{3x}-4}{\text{7}}+\frac{\text{7}}{\text{3x}-4}=\frac{5}{2},$ $\text{x}\neq\frac{4}{3}$
Answer$\frac{\text{3x}-4}{\text{7}}+\frac{\text{7}}{\text{3x}-4}=\frac{5}{2}$
Taking $\frac{\text{3x}-4}{7}=\text{y},$ we have
$\text{y}+\frac{1}{\text{y}}=\frac{5}{2}$
$\Rightarrow\frac{\text{y}^2+1}{\text{y}}=\frac{5}{2}$
$\Rightarrow 2x^2 + 2 = 5y = 0$
$\Rightarrow 2y^2 - 5y + 2 = 0$
$\Rightarrow 2y^2 - 4y - y + 2 = 0$
$\Rightarrow 2y(y - 2) - 1(y - 2) = 0$
$\Rightarrow (y - 2)(2y - 1) = 0$
$\Rightarrow y - 2 = 0 or 2y - 1 = 0$
⇒ y = 2 or $\text{y}=\frac{\text{1}}{2}$
$\Rightarrow\frac{\text{3x}-4}{7}=2$ or $\frac{\text{3x}-4}{7}=\frac{1}{2}$
⇒ 3x - 4 = 14 or 6x - 8 = 7
⇒ 3x = 18 or 6x = 15
⇒ x = 6 or $\text{x}=\frac{5}{2}$
View full question & answer→Question 1165 Marks
Solve the following quadratic equation:
$3\Big(\frac{\text{3x}-1}{\text{2x}+3}\Big)-2\Big(\frac{\text{2x}+3}{\text{3x}-1}\Big)=5$ $\text{x}\neq\frac{1}{3},\ \frac{-3}{2}$
Answer$3\Big(\frac{\text{3x}-1}{\text{2x}+3}\Big)-2\Big(\frac{\text{2x}+3}{\text{3x}-1}\Big)=5$
Taking $\frac{\text{3x}-1}{\text{2x}+3}=\text{y},$ we have
$\text{3y}-\frac{2}{\text{y}}=5$
$\Rightarrow\frac{\text{3y}^2-2}{\text{y}}=5$
$\Rightarrow 3y^2 - 2 = 5y$
$\Rightarrow 3y^2 - 5y - 2 = 0$
$\Rightarrow 3y^2 - 6y + y - 2 = 0$
$\Rightarrow 3y(y - 2) + 1(y - 2) = 0$
$\Rightarrow (y - 2)(3y + 1) = 0$
$\Rightarrow y - 2 = 0 or 3y + 1 = 0$
⇒ y = 2 or $\text{y}=\frac{-1}{3}$
$\Rightarrow\frac{\text{3x}-1}{\text{2x}+3}=2$ or $\frac{\text{3x}-1}{\text{2x}+3}=\frac{-1}{3}$
$\Rightarrow 3x - 1 = 4x + 6 or 9x - 3 = -2x - 3$
$\Rightarrow x = -7 or 11x = 0$
$\Rightarrow x = -7 or x = 0$
View full question & answer→Question 1175 Marks
Solve the following quadratic equation:
$\frac{\text{1}}{\text{x}+1}+\frac{\text{3}}{\text{5x}+1}=\frac{5}{\text{x}+4},$ $\text{x}\neq-1,-\frac{1}{5},-4$
Answer$\frac{\text{1}}{\text{x}+1}+\frac{\text{3}}{\text{5x}+1}=\frac{5}{\text{x}+4}\ \dots(1)$
Equation (1):
$\frac{\text{1}}{\text{x}+1}+\frac{\text{3}}{\text{5x}+1}=\frac{5}{\text{x}+4}$
Taking LCM
$\frac{(\text{5x}+1)+3(\text{x}+1)}{(\text{x}+1)(\text{5x}+1)}=\frac{5}{\text{x}+4}$
$\frac{\text{5x}+1+3\text{x}+3}{(\text{5x}^2+\text{6x}+1)}=\frac{5}{\text{x}+4}$
$\frac{\text{8x}+4}{(\text{5x}^2+\text{6x}+1)}=\frac{5}{\text{x}+4}$
Cross multiply
$\Rightarrow (8x + 4)(x + 4) = (5x^2 + 6x + 1)5$
$\Rightarrow 8x^2 - 25x^2 + 36x - 30x + 16 - 5 = 0$
$\Rightarrow -17x^2 + 6x + 11 = 0$
$\Rightarrow 17x^2 - 6x - 11 = 0$
$\Rightarrow 17x^2 - 17x + 11x - 11 = 0$ (Middle term split)
⇒ (x - 1)(17x + 11) = 0
⇒ x = 1 or $\text{x}=\frac{-11}{17}$
View full question & answer→Question 1185 Marks
Solve the following quadratic equation:
$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7},$ $\text{x}\neq1,\ -5$
Answer$\frac{1}{\text{x}-1}-\frac{1}{\text{x}+5}=\frac{6}{7}$
$\Rightarrow\frac{\text{x}+5-\text{x}+1}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{6}{\text{x}^2+\text{5x}-\text{x}-5}=\frac{6}{7}$
$\Rightarrow\frac{1}{\text{x}^2+\text{4x}-5}=\frac{1}{7}$
$\Rightarrow 7 = x^2 + 4x - 5$
$\Rightarrow x^2 + 4x - 12 = 0$
$\Rightarrow x^2 + 6x - 2x - 12 = 0$
$\Rightarrow x(x + 6) - 2(x + 6) = 0$
$\Rightarrow (x + 6)(x - 2) = 0$
$\Rightarrow x + 6 = 0 or x - 2 = 0$
$\Rightarrow x = -6 or x = 2$
View full question & answer→Question 1195 Marks
Solve the following quadratic equation:
$\frac{\text{x}}{\text{x}+1}+\frac{\text{x}+1}{\text{x}}=2\frac{4}{15},$ $\text{x}\neq0,-1$
Answer$\frac{\text{x}}{\text{x}+1}+\frac{\text{x}+1}{\text{x}}=2\frac{4}{15}$
$\Rightarrow\frac{\text{x}}{\text{x}+1}+\frac{\text{x}+1}{\text{x}}=\frac{34}{15}$
Taking $\frac{\text{x}}{\text{x}+1}=\text{y},$ we have
$\text{y}+\frac{1}{\text{y}}=\frac{34}{15}$
$\Rightarrow\frac{\text{y}^2+1}{\text{y}}=\frac{34}{15}$
$\Rightarrow 15y^2 + 15 = 34y = 0$
$\Rightarrow 15y^2 - 34y + 15 = 0$
$\Rightarrow 15y^2 - 25y - 9y + 15 = 0$
$\Rightarrow 5y(3y - 5) - 3(3y - 5) = 0$
$\Rightarrow (3y - 5)(5y - 3) = 0$
$\Rightarrow 3y - 5 = 0 or 5y - 3 = 0$
$\Rightarrow\text{y}=\frac{5}{3}$ or $\text{y}=\frac{\text{3}}{5}$
$\Rightarrow\frac{\text{x}}{\text{x}+1}=\frac{5}{3}$ or $\frac{\text{x}}{\text{x}+1}=\frac{3}{5}$
⇒ 3x = 5x + 5 or 5x = 3x + 3
⇒ 2x = -5 or 2x = 3
$\Rightarrow\text{x}=\frac{-5}{2}$ or $\text{x}=\frac{3}{2}$
View full question & answer→Question 1205 Marks
Solve the following quadratic equation:
$4^{(x+1)} + 4^{(1-x)} = 10$
Answer$4^{(x+1)} + 4^{(1-x)} = 10$
$4^x.4^1 + 4^1.4^{-x} = 10$
$\Rightarrow\text{4y}+\frac{4}{\text{y}}=10$ where $4^x = y$
$\Rightarrow 4y^2 - 10y + 4 = 0$
$\Rightarrow 4y^2 - 8y - 2y + 4 = 0$
$\Rightarrow 4y(y - 2) - 2(y - 2) = 0$
$\Rightarrow (y - 2)(4y - 2) = 0$
⇒ y - 2 = 0 or 4y - 2 = 0
⇒ y = 2 or $\text{y}=\frac{2}{4}=\frac{1}2{}$
⇒ y = 2 or $\text{y}=\frac{1}{2}$
In case I:
$\Rightarrow 4^x = 2$
$\Rightarrow (2)^{2x} = (2)^1$
$\Rightarrow 2x = 1$
$\Rightarrow\text{x}=\frac{1}{2}$
In case II:
$\Rightarrow4^\text{x}=\frac{1}{2}$
$\Rightarrow(2)^\text{2x}=\Big(\frac{1}{2}\Big)^1$
$\Rightarrow(2)^{\text{2x}}=(2)^{-1}$
$\Rightarrow\text{x}=-\frac{1}{2}$
Hence, $\frac{1}{2},-\frac{1}2{}$ are the roots of given equation.
View full question & answer→Question 1215 Marks
Solve the following quadratic equation:
$\frac{\text{a}}{(\text{x}-\text{b})}+\frac{\text{b}}{(\text{x}-\text{a})}=2,$ $\text{x}\neq-\text{b},\ \text{a}$
AnswerThe given equation:
$\Big(\frac{\text{a}}{\text{x}-\text{b}}-1\Big)+\Big(\frac{\text{b}}{\text{x}-\text{a}}-1\Big)=0$
$\Rightarrow\frac{(\text{a}-\text{x}+\text{b})}{(\text{x}-\text{b})}+\frac{(\text{b}-\text{x}+\text{a})}{(\text{x}-\text{a})}=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})\Big[\frac{1}{(\text{x}-\text{b})}+\frac{1}{(\text{x}-\text{a})}\Big]=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})\Big[\frac{\text{2x}-(\text{a}+\text{b})}{(\text{x}-\text{a})(\text{x}-\text{b})}\Big]=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})[\text{2x}-(\text{a}+\text{b})]=0$
$\Rightarrow\text{x}=(\text{a}+\text{b})$ or $\text{x}=\frac{(\text{a}+\text{b})}{2}$
Hence, (a + b) and $\frac{(\text{a}+\text{b})}{2}$ is the root of the given equation.
View full question & answer→Question 1225 Marks
Solve the following quadratic equation:
$\frac{\text{x}}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}=4\frac{1}{4},$ $\text{x}\neq0,1$
Answer$\frac{\text{x}}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}=4\frac{1}{4}$
$\Rightarrow\frac{\text{x}}{\text{x}-1}+\frac{\text{x}-1}{\text{x}}=\frac{17}{4}$
Taking $\frac{\text{x}}{\text{x}-1}=\text{y},$ we have
$\text{y}+\frac{1}{\text{y}}=\frac{17}{4}$
$\Rightarrow\frac{\text{y}^2+1}{\text{y}}=\frac{17}{4}$
$\Rightarrow 4y^2 + 4 = 17y = 0$
$\Rightarrow 4y^2 - 17y + 4 = 0$
$\Rightarrow 4y^2 - 16y - y + 4 = 0$
$\Rightarrow 4y(y - 4) - 1(y - 4) = 0$
$\Rightarrow (y - 4)(4y - 1) = 0$
$\Rightarrow y - 4 = 0 or 4y - 1 = 0$
⇒ y = 4 or $\text{y}=\frac{\text{1}}{4}$
$\Rightarrow\frac{\text{x}}{\text{x}-1}=4$ or $\frac{\text{x}}{\text{x}-1}=\frac{1}{4}$
⇒ x = 4x - 4 or 4x = x - 1
⇒ 3x = 4 or 3x = -1
$\Rightarrow\text{x}=\frac{4}{3}$ or $\text{x}=\frac{-1}{3}$
View full question & answer→Question 1235 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$x^2 + 5x - (a^2 + a - 6) = 0$
AnswerThe given equation is $x^2+5 x-\left(a^2+a-6\right)=0$.
Comparing it with $Ax ^2+ Bx + C =0$, we get $A =1, B=5$ and $C =-\left( a ^2+ a -6\right)$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= 5^2 - 4 \times 1 \times [-(a^2 + a - 6)] = 25 + 4a^2 + 4a - 24 = 4a^2 + 4a + 1 = (2a + 1)^2 > 0$ So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{(2\text{a}+1)^2}=2\text{a}+1$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-5+2\text{a}+1}{2\times1}$
$=\frac{2\text{a}-4}{2}$
$=\text{a}-2$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-5-(2\text{a}+1)}{2\times1}$
$=\frac{-2\text{a}-6}{2}$
$=-\text{a}-3$
$=-(\text{a}+3)$
Hence, (a - 2) and -(a + 3) are the roots of the given equation.
View full question & answer→Question 1245 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 2ax - (4b^2 - a^2) = 0$
AnswerThe given equation is $x^2-2 a x-\left(4 b^2-a^2\right)=0$. Comparing it with $A x^2+B x+C=0$, we get $A=1, B=-2 a$ and $C=-$ $\left(4 b^2-a^2\right) \therefore$ Discrimiant, $D=B^2-4 A C$ $=(-2 a)^2-4 \times 1 \times\left[-\left(4 b^2-a^2\right)\right]=4 a^2+16 b^2-4 a^2=16 b^2>0$ So, the given equation has real roots.Now, $\sqrt{\text{D}}=\sqrt{16\text{b}^2}=4\text{b}$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-2\text{a})+4\text{b}}{2\times1}$
$=\frac{2(\text{a}+2\text{b})}{2}$
$=\text{a}+2\text{b}$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-(-2\text{a}-4\text{b})}{2\times1}$
$=\frac{2(\text{a}-2\text{b})}{2}$
$=\text{a}-2\text{b}$
Hence, a + 2b and a - 2b are the roots of the given equation.
View full question & answer→Question 1255 Marks
Solve the following quadratic equation:
$\Big(\frac{\text{x}}{\text{x}}+1\Big)^2-5\Big(\frac{\text{x}}{\text{x}+1}\Big)+6=0,$ $\text{x}\neq-1$
AnswerPutting $\Big(\frac{\text{x}}{\text{x}+1}\Big)=\text{y},$ the given equation become.
$\Rightarrow y^2 - 5y + 6 = 0$
$\Rightarrow y^2 - 3y - 2y + 6 = 0$
$\Rightarrow y(y - 3) - 2(y - 3) = 0$
$\Rightarrow (y - 3)(y - 2) = 0$
$\Rightarrow y - 3 = 0 or y - 2 = 0$
$\Rightarrow y = 3 or y = 2$
Case I:y = 3
$\Rightarrow\frac{\text{x}}{\text{x}+1}=3 $
$\Rightarrow 3x + 3 = x$
$\Rightarrow 3x - x = 3$
$\Rightarrow 2x = 3$
$\Rightarrow\text{x}=\frac{-3}{2}$
Case II:y = 2
$\Rightarrow\frac{\text{x}}{\text{x}+1}=2 $
$\Rightarrow 2x + 2 = x$
$\Rightarrow 2x - x = -2$
$\Rightarrow x = -2$ Hence,
$\frac{-3}{2},\ -2$ are the roots of the given equation.
View full question & answer→Question 1265 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula$:x^2 - (2b - 1)x + (b^2 - b - 20) = 0$
AnswerThe given equation is $x^2-(2 b-1) x+\left(b^2-b-20\right)=0$. Comparing it with $A x^2+B x+C=0$, we get $A=1, B=-(2 b-1)$ and $C=b^2-b-20$
$\therefore$ Discriminant, $D = B^2 - 4AC$
$= [-(2b - 1)]^2 - 4 \times 1 \times (b^2 - b - 20)$
$= 4b^2 - 4b + 1 - 4b^2 + 4b + 80$
$= 81 > 0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{81}=9$
$\therefore\alpha=\frac{-\text{B}+\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(2\text{b}-1)]+9}{2\times1}$
$=\frac{2\text{b}+8}{2}$
$=\text{b}+4$
$\beta=\frac{-\text{B}-\sqrt{\text{D}}}{2\text{A}}$
$=\frac{-[-(2\text{b}-1)]-9}{2\times1}$
$=\frac{2\text{b}-10}{2}$
$=\text{b}-5$
Hence, (b + 4) and (b - 5) are the roots of the given equation.
View full question & answer→Question 1275 Marks
Solve the following quadratic equation:$12abx^2 - (9a^2 - 8b^2)x - 6ab = 0$
Answer$12abx^2 - (9a^2 - 8b^2)x - 6ab = 0$
$\Rightarrow 12abx^2 - 9a^2x + 8b^2x - 6ab = 0$
$\Rightarrow 3ax(4bx - 3a) + 2b(4bx - 3a) = 0$
$\Rightarrow (4bx - 3a)(3ax + 2b) = 0$
$\Rightarrow (4bx - 3a) = 0 or (3ax + 2b) = 0$
$\Rightarrow 4bx = 3a or 3ax = -2b$
$\Rightarrow\text{x}=\frac{\text{3a}}{\text{4b}}$ or $\text{x}=\frac{-\text{2b}^2}{\text{3a}}$
Hence, $\frac{\text{3a}}{\text{4b}}$ and $\frac{-\text{2b}}{\text{3a}}$ are the roots of the given equation.
View full question & answer→Question 1285 Marks
Solve the following quadratic equation:
$\frac{\text{x}+3}{\text{x}-2}-\frac{1-\text{x}}{\text{x}}=4\frac{1}{4},$ $\text{x}\neq2,\ 0$
Answer$\frac{\text{x}+3}{\text{x}-2}-\frac{1-\text{x}}{\text{x}}=4\frac{1}{4}$
$\Rightarrow\frac{\text{x}(\text{x}+3)-(1-\text{x})(\text{x}-2)}{\text{x}(\text{x}-2)}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}^2+\text{3x}-(\text{x}-2-\text{x}^2+\text{2x})}{\text{x}^2-\text{2x}}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}^2+\text{3x}-(\text{3x}-2-\text{x}^2)}{\text{x}^2-\text{2x}}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}^2+\text{3x}-\text{3x}+2+\text{x}^2}{\text{x}^2-\text{2x}}=\frac{17}{4}$
$\Rightarrow\frac{\text{2x}^2+2}{\text{x}^2-\text{2x}}=\frac{17}{4}$
$\Rightarrow 8x^2 + 8 = 17x^2 - 34x = 0$
$\Rightarrow 9x^2 - 34x - 8 = 0$
$\Rightarrow 9x^2 - 36x + 2x - 8 = 0$
$\Rightarrow 9x(x - 4) + 2(x - 4) = 0$
$\Rightarrow (x - 4)(9x + 2) = 0$
$\Rightarrow x - 4 = 0 or 9x + 2 = 0$
$\Rightarrow x = 4$ or $\text{x}=\frac{-\text{2}}{9}$
View full question & answer→Question 1295 Marks
Find the values of k for which the given quadratic equation has real and distinct roots:
$9x^2 + 3kx + 4 = 0$
AnswerThe given equation is$ 9x^2 + 3kx + 4 = 0$
$\therefore$$ D = (3k)^2 - 4 \times 9 \times 4$
$D = 9k^2 - 144$
The given equation has real and distinct roots if D > 0.
$\therefore$ $9k^2 - 144 > 0$
$\Rightarrow 9(k^2 - 16) > 0$
$\Rightarrow (k - 4)(k + 4) > 0$
$\Rightarrow k < -4 or k > 4$
View full question & answer→Question 1305 Marks
Solve the following quadratic equation:
$\frac{1}{\text{2a}+\text{b}+\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}+\frac{1}{\text{2x}}$
Answer$\frac{1}{\text{2a}+\text{b}+\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}+\frac{1}{\text{2x}}$
$\Rightarrow\frac{1}{\text{2a}+\text{b}+\text{2x}}-\frac{1}{\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}$
$\Rightarrow\frac{\text{2x}-(\text{2a}+\text{b}+\text{2x})}{\text{2x}(\text{2a}+\text{b}+\text{2x})}=\frac{\text{b}+\text{2a}}{\text{2ab}{}}$
$\Rightarrow\frac{\text{2x}-\text{2a}-\text{b}-\text{2x}}{\text{4ax}+\text{2bx}+\text{4x}^2}=\frac{\text{b}+\text{2a}}{\text{2ab}{}}$
$\Rightarrow\frac{-(\text{2a}+\text{b})}{\text{2x}(\text{2a}+\text{b}+\text{2x})}=\frac{\text{b}+\text{2a}}{\text{2ab}}$
$\Rightarrow\frac{-1}{\text{x}(\text{2a}+\text{b}+\text{2x})}=\frac{1}{\text{ab}}$
$\Rightarrow -ab = 2ax + bx + 2x^2$
$\Rightarrow 2x^2 + bx + 2ax + ab = 0$
$\Rightarrow 2x^2 + 2ax + bx + ab = 0$
$\Rightarrow 2x(x + a) + b(x + a) = 0$
$\Rightarrow (x + a)(2x + b) = 0$
$\Rightarrow x + a = 0 or 2x + b = 0$
⇒ x = -a or $\text{x}=\frac{-\text{b}}{2}$
View full question & answer→Question 1315 Marks
Solve the following equations by using the method of completing the square:
$7x^2 + 3x - 4 = 0$
Answer$7x^2 + 3x - 4 = 0$
$\Rightarrow 49x^2 + 21x - 28 = 0$ (Multiplying both sides by 7)
$\Rightarrow 49x^2 + 21x = 28$
$\Rightarrow(\text{7x})^2+2\times\text{7x}\times\frac{3}{2}+\Big(\frac{3}{2}\Big)^2\\=28+\Big(\frac{3}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{3}2{}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{7x}+\frac{3}{2}\Big)^2$
$=29+\frac{9}{4}$
$=\frac{121}{4}=\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{7x}-\frac{3}{2}=\pm\frac{11}{2}$ (Taking square root on both sides)
$\Rightarrow\text{7x}+\frac{3}{2}=\frac{11}{2}$ or $\text{7x}+\frac{3}{2}=-\frac{11}{2}$
$\Rightarrow\text{7x}=\frac{11}{2}-\frac{3}{2}=\frac{8}{2}=4$ or $\text{7x}=-\frac{11}{2}-\frac{3}{2}=-\frac{14}{2}=-7$
$\Rightarrow\text{x}=\frac{4}{7}$ or x = -1
Hence, $\frac{4}{7}$ and -1 are the roots of the given equation.
View full question & answer→Question 1325 Marks
Solve the following quadratic equation:
$\frac{1}{\text{x}-2}+\frac{2}{\text{x}-1}=\frac{6}{\text{x}}$ $\text{x}\neq0,\ 1,\ 2$
Answer$\frac{1}{\text{x}-2}+\frac{2}{\text{x}-1}=\frac{6}{\text{x}}$ $\Rightarrow\frac{\text{x}-1+\text{2x}-4}{(\text{x}-2)(\text{x}-1)}=\frac{6}{\text{x}}$ $\Rightarrow\frac{\text{3x}-5}{\text{x}^2-\text{x}-\text{2x}+2}=\frac{6}{\text{x}}$ $\Rightarrow\frac{\text{3x}-5}{\text{x}^2-\text{3x}+2}=\frac{6}{\text{x}}$
$\Rightarrow 3x^2 - 5x = 6x^2 - 18x + 12$
$\Rightarrow 3x^2 - 13 + 12 = 0$
$\Rightarrow 3x^2 - 9x - 4x + 12 = 0$
$\Rightarrow 3x(x - 3) - 4(x - 3) = 0$
$\Rightarrow (x - 3)(3x - 4) = 0$
$\Rightarrow x - 3 = 0 or 3x - 4 = 0$
⇒ x = 3 or $\text{x}=\frac{4}{3}$
View full question & answer→Question 1335 Marks
Solve the following quadratic equation:
$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{\text{3x}-1},$ $\text{x}\neq-1,\ \frac{1}{3}$
Answer$\frac{3}{\text{x}+1}-\frac{1}{2}=\frac{2}{\text{3x}-1}$
$\Rightarrow\frac{3}{\text{x}+1}-\frac{2}{\text{3x}-1}=\frac{1}{2}$
$\Rightarrow\frac{3(\text{3x}-1)-2(\text{x}+1)}{(\text{x}+1)(\text{3x}-1)}=\frac{1}{2}$
$\Rightarrow\frac{\text{9x}-3-\text{2x}-2}{\text{3x}^2-\text{x}+\text{3x}-1}=\frac{1}{2}$
$\Rightarrow\frac{\text{7x}-5}{\text{3x}^2+\text{2x}-1}=\frac{1}{2}$
$\Rightarrow 14x - 10 = 3x^2 + 2x - 1$
$\Rightarrow 3x^2 + 2x - 1 - 14x + 10 = 0$
$\Rightarrow 3x^2 - 12x + 9 = 0$
$\Rightarrow x^2 - 4x + 3 = 0$
$\Rightarrow x^2 - 3x - x + 3 = 0$
$\Rightarrow x(x - 3) - 1(x - 3) = 0$
$\Rightarrow (x - 3)(x - 1) = 0$
$\Rightarrow x - 3 = 0 or x - 1 = 0$
$\Rightarrow x = 3 or x = 1$
View full question & answer→Question 1345 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:$2\text{x}^2+5\sqrt3\text{x}+6=0$
AnswerThe given equation is $2\text{x}^2+5\sqrt3\text{x}+6=0$Comparing it with $ax^2 + bx + c = 0$, we get
$\text{a}=2,\ \text{b}=5\sqrt3$ and $\text{c}=6$
$\therefore$ Discriminant, $\text{D}=\text{b}^2-4\text{ac}$
$=\big(5\sqrt3\big)^2-4\times2\times6$
$=75-48$
$=27>0$
So, the given equation has real roots.
Now, $\sqrt{\text{D}}=\sqrt{27}=3\sqrt3$
$\therefore\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5\sqrt3+3\sqrt3}{2\times2}$
$=\frac{-2\sqrt3}{4}$
$=-\frac{\sqrt3}{2}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-5\sqrt3-3\sqrt3}{2\times2}$
$=\frac{-8\sqrt3}{4}$
$=-2\sqrt3$
Hence, $-\frac{\sqrt3}{2}$ and $-2\sqrt3$ are the roots of the given equation.
View full question & answer→Question 1355 Marks
Solve the following quadratic equation:
$\frac{\text{a}}{(\text{ax}-\text{1})}+\frac{\text{b}}{(\text{bx}-\text{1})}=(\text{a}+\text{b}),$ $\text{x}\neq\frac{1}{\text{a}},\ \frac{1}{\text{b}}$
Answer$\frac{\text{a}}{(\text{ax}-\text{1})}+\frac{\text{b}}{(\text{bx}-\text{1})}=(\text{a}+\text{b})$
$\Rightarrow\Big[\frac{\text{a}}{\text{ax}-1}-\text{b}\Big]+\Big[\frac{\text{b}}{\text{bx}-1}-\text{a}\Big]=0$
$\Rightarrow\Big[\frac{\text{a}-\text{abx}+\text{b}}{\text{ax}-1}\Big]+\Big[\frac{\text{b}-\text{abx}+\text{a}}{\text{bx}-1}\Big]=0$
$\Rightarrow(\text{a}-\text{abx}+\text{b})\Big[\frac{1}{\text{ax}-1}+\frac{1}{\text{bx}-1}\Big]=0$
$\Rightarrow(\text{a}-\text{abx}+\text{b})$ or $\frac{1}{\text{ax}-1}+\frac{1}{\text{bx}-1}=0$
$\Rightarrow\text{abx}=\text{a}+\text{b}$ or $\frac{1}{\text{ax}-1}=-\frac{1}{\text{bx}-1}$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{bx} - 1 = -\text{ax} + 1$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{bx} + \text{ax} = 2$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{x}(\text{b} + \text{a}) = 2$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{x}=\frac{2}{\text{a}+\text{b}}$
View full question & answer→Question 1365 Marks
Solve the following quadratic equation:
$\frac{\text{1}}{\text{x}+1}+\frac{\text{2}}{\text{x}+2}=\frac{5}{\text{x}+4},$ $\text{x}\neq-1,-2,-4$
Answer$\frac{\text{1}}{\text{x}+1}+\frac{\text{2}}{\text{x}+2}=\frac{5}{\text{x}+4}$
$\Rightarrow\frac{\text{x}+2+\text{2x}+2}{(\text{x}+1)(\text{x}+2)}=\frac{5}{\text{x}+4}$
$\Rightarrow\frac{\text{3x}+4}{\text{x}^2+\text{3x}+2}=\frac{5}{\text{x}+4}$
$\Rightarrow (3x + 4)(x + 4) = 5(x^2 + 3x + 2)$
$\Rightarrow 3x^2 + 16x + 16 = 5x^2 + 15x + 10$
$\Rightarrow 2x^2 - x - 6 = 0$
$\Rightarrow 2x^2 - 4x + 3x - 6 = 0$
$\Rightarrow 2x(x - 2) + 3(x - 2) = 0$
$\Rightarrow (x - 2)(2x + 3) = 0$
$\Rightarrow x - 2 = 0 or 2x + 3 = 0$
⇒ x = 2 or $\text{x}=\frac{-3}{2}$
View full question & answer→Question 1375 Marks
Solve the following quadratic equation:
$3^{(x+2)} + 3^{-x} = 10$
Answer$3^{(x+2)} + 3^{-x} = 10$
$3^x.3^2 + 3^{-x} = 10$
$\Rightarrow\text{9y}+\frac{1}{\text{y}}=10$ where $3^x = y$
$\Rightarrow 9y^2 - 10y + 1 = 0$
$\Rightarrow 9y^2 - 9y - y + 1 = 0$
$\Rightarrow 9y(y - 1) - 1(y - 1) = 0$
$\Rightarrow (9y - 1)(y - 1) = 0$
$\Rightarrow 9y - 1 = 0 or y - 1 = 0$
$\Rightarrow\text{y}=\frac{1}{9}$ or y = 1
If $3^\text{x}=\frac{1}{9}$
$\Rightarrow 3x = (3)^{-2}$
$\Rightarrow x = -2$
$If 3x = 1 = 30$
$\Rightarrow x = 0$
Hence, -2, 0 are the roots of given equation.
View full question & answer→Question 1385 Marks
Solve the following equations by using the method of completing the square:
$\sqrt2\text{x}^2-3\text{x}-2\sqrt2=0$
Answer$\sqrt2\text{x}^2-3\text{x}-2\sqrt2=0$
$\Rightarrow2\text{x}^2-3\sqrt2\text{x}-4=0$ $\big($Multiplying both sides by $\sqrt2\big)$
$\Rightarrow\text{2x}^2-3\sqrt2\text{x}=4$
$\Rightarrow\big(\sqrt2\text{x}\big)^2-2\times\sqrt2\text{x}\times\frac{3}2{}+\Big(\frac{3}{2}\Big)^2\\=4+\Big(\frac{3}{2}\Big)^2$ $\Big[$Adding $\Big(\frac{3}{2}\Big)^2$ on both sides$\Big]$
$\Rightarrow\Big(\text{2x}-\frac{3}{2}\Big)^2=4+\frac{9}{4}$
$=\frac{25}{4}=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\sqrt2\text{x}-\frac{3}{2}=\pm\frac{5}{2}$ (Taking square root on both sides)
$\Rightarrow\sqrt2\text{x}-\frac{3}{2}=\frac{5}{2}$ or $\sqrt2\text{x}-\frac{3}{2}=-\frac{5}{2}$
$\Rightarrow\sqrt2\text{x}=\frac{5}2{}+\frac{3}{2}=\frac{8}{2}=4$ or $\sqrt2\text{x}=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1$
$\Rightarrow\text{x}=\frac{4}{\sqrt2}=2\sqrt2$ or $\text{x}=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2}$
Hence, $2\sqrt2$ and $-\frac{\sqrt2}{2}$ are the roots of the given equation.
View full question & answer→Question 1395 Marks
Solve the following quadratic equation:$a^2b^2x^2 + b^2x - a^2x - 1= 0$
Answer$a^2b^2x^2 + b^2x - a^2x - 1= 0\Rightarrow b^2x(a^2x + 1) - 1(a^2x + 1) = 0$
$\Rightarrow (a^2x + 1)(b^2x - 1) = 0$
$\Rightarrow (a^2x + 1) = 0 or (b^2x - 1) = 0$
$\Rightarrow\text{x}=\frac{-\text{1}}{\text{a}^2}$ or $\text{x}=\frac{\text{1}}{\text{b}^2}$
Hence, $\frac{-\text{1}}{\text{a}^2}$ and $\frac{\text{1}}{\text{b}^2}$ are the roots of the given equation.
View full question & answer→Question 1405 Marks
Find the roots of the following equation, if they exist, by applying the quadratic formula:
$3a^2x^2 + 8abx + 4b^2 = 0$, $\text{a}\neq0$
AnswerGiven,
$3a^2x^2 + 8abx + 4b^2 = 0$
On comparing it with $Ax^2 + Bx + C = 0$, we get:
$A = 3a^2, B = 8ab$ and$ C = 4b^2$
Discriminant D is given by:
$D = (B^2 - 4AC)$
$= (8ab)^2 - 4 \times 3a^2 \times 4b^2$
$= 16a^2b^2 > 0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by:
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-8\text{ab}+\sqrt{16\text{a}^2\text{b}^2}}{2\times3\text{a}^2}$
$=\frac{-8\text{ab}+4\text{ab}}{6\text{a}^2}$
$=\frac{-4\text{ab}}{6\text{a}^2}$
$=\frac{-2\text{b}}{3\text{a}}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-8\text{ab}-\sqrt{16\text{a}^2\text{b}^2}}{2\times3\text{a}^2}$
$=\frac{-8\text{ab}-4\text{ab}}{6\text{a}^2}$
$=\frac{-12\text{ab}}{6\text{a}^2}$
$=\frac{-2\text{b}}{\text{a}}$
Thus, the roots of the equation are $\frac{-2\text{b}}{3\text{a}}$ and $\frac{-2\text{b}}{\text{a}}.$
View full question & answer→Question 1415 Marks
Solve the following quadratic equation:
$3\Big(\frac{\text{7x}+1}{\text{5x}-3}\Big)-4\Big(\frac{\text{5x}-3}{\text{7x}+1}\Big)=11$ $\text{x}\neq\frac{3}{5},\ \frac{-1}{7}$
Answer$3\Big(\frac{\text{7x}+1}{\text{5x}-3}\Big)-4\Big(\frac{\text{5x}-3}{\text{7x}+1}\Big)=11$
Taking $\frac{\text{7x}+1}{\text{5x}-3}=\text{y},$ we have
$\text{3y}-\frac{4}{\text{y}}=11$
$\Rightarrow\frac{\text{3y}^2-4}{\text{y}}=11$
$\Rightarrow 3y^2 - 4 = 11y$
$\Rightarrow 3y^2 - 11y - 4 = 0$
$\Rightarrow 3y^2 - 12y + y - 4 = 0$
$\Rightarrow 3y(y - 4) + 1(y - 4) = 0$
$\Rightarrow (y - 4)(3y + 1) = 0$
$\Rightarrow y - 4 = 0 or 3y + 1 = 0$
⇒ y = 4 or $\text{y}=\frac{-1}{3}$
$\Rightarrow\frac{\text{7x}+1}{\text{5x}-3}=4$ or $\frac{\text{7x}+1}{\text{5x}-3}=\frac{-1}{3}$
$\Rightarrow 7x + 1 = 20x - 12 or 21x + 3 = -5x + 3$
$\Rightarrow 13x = 13 or 26x = 0$
$\Rightarrow x = 1 or x = 0$
View full question & answer→Question 1425 Marks
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$3x^2 - 2x + 2 = 0$
AnswerThe given equation is $3x^2 - 2x + 2 = 0$ C0mparing it with $ax^2 + bx + c = 0$, we get a = 3, b = -2 and c = 2$\therefore$ Discriminant,$ D = b^2 - 4ac$
$= (-2)^2 - 4 \times 3 \times 2 = 4 - 24 = -20 < 0$ Hence, the given equation has no real roots (or real roots does not exist).
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Solve the following quadratic equation:$4x^2 - 2(a^2 + b^2)x + a^2b^2 = 0$
Answer$4x^2 - 2(a^2 + b^2)x + a^2b^2 = 0$
$\Rightarrow 4x^2 - 2a^2x - 2b^2x + a^2b^2 = 0$
$\Rightarrow 2x(2x - a^2) - b^2(2x - a^2) = 0$
$\Rightarrow (2x - a^2)(2x - b^2) = 0$
$\Rightarrow (2x - a^2) = 0 or (2x - b^2) = 0$
$\Rightarrow\text{x}=\frac{\text{a}^2}{\text{2}}$ or $\text{x}=\frac{\text{b}^2}{\text{2}}$
Hence, $\frac{\text{a}^2}{\text{2}}$ and $\frac{\text{b}^2}{\text{2}}$ are the roots of the given equation.
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Solve the following quadratic equation:
$\frac{\text{x}-1}{\text{2x}+1}+\frac{\text{2x}+1}{\text{x}-1}=2,$ $\text{x}\neq-\frac{1}{2},\ 1$
Answer$\frac{\text{x}-1}{\text{2x}+1}+\frac{\text{2x}+1}{\text{x}-1}=2$
$\frac{(\text{x}-1)^2+(\text{2x}+1)^2}{(\text{x}-1)(\text{2x}+1)}=2$
$\Rightarrow x^2 - 2x + 1 + 4x^2 + 4x + 1 = 2(x - 1)(2x + 1)$
$\Rightarrow 5x^2 + 2x + 2 = 4x^2 - 2x - 2$
$\Rightarrow x^2 + 4x + 4 = 0$
$\Rightarrow (x + 2)^2 = 0$
$\Rightarrow x = -2, 2$
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