Question 513 Marks
Find the value of k for which the following equations have real and equal roots:
$(k+1) x^2-2(k-1) x+1=0$
Answer$(k+1) x^2-2(k-1) x+1=0$
Here $a=k+1, b=-2(k-1)$ and $c=1$
$ \therefore \text { Discriminant }(D)=b^2-4 a c $
$ =[-2(k-1)]^2-4(k+1) \times 1 $
$ =4\left(k^2-2 k+1\right)-4(k+1) $
$ =4 k^2-8 k+4-4 k-4 $
$ =4 k^2-12 k$
$\because$ Roots are real and equal
$ \therefore D=0 $
$ 4 k^2-12 k=0 $
$ k^2-3 k=0(\text { Dividing by } 4) $
$ k(k-3)=0$
Either $k=0$
or $k-3=0$, then $k=3$
$\therefore \mathrm{k}=0,3$
View full question & answer→Question 523 Marks
Find the value of k for which the root are real and equal in the following equations:
$k x^2+4 x+1=0$
Answer$k x^2+4 x+1=0$
Here $a=k, b=4, c=1$
$ \therefore \text { Discriminant }(D)=b^2-4 a c $
$=(4)^2-4 \times \mathrm{k} \times 1$
$ =16-4 \mathrm{k}$
$\because$ Roots are real and equal
$\therefore \mathrm{D}=0$
$⇒ 16 - 4k = 0$
$⇒ 4k = 16$
$\Rightarrow\text{k}=\frac{16}{4}=4$
Hence, $k = 4$
View full question & answer→Question 533 Marks
The sum of a number and its reciprocal is $\frac{17}{4}.$ Find the number.
AnswerLet a number be x abd its reciprocal is $\frac{1}{\text{x}}$
Then according to question
$\text{x}+\frac{1}{\text{x}}=\frac{17}{4}$
$\frac{\text{x}^2+1}{\text{x}}=\frac{17}{4}$
By cross multoplication,
$4 x^2+4=17 x$
$4 x^2-17 x+4=0$
$4 x^2-17 x+4=0$
$4 x^2-x-16 x+4=0$
$x(4 x-1)-4(4 x-1)=0$
$(4 x-1)(x-4)=0$
$(4 x-1)=0$
$x=\frac{1}{4} $
Or $(x-4)=0 $
$ x=4$
Thus, two consecutive number be either 4 or $\frac{1}{4}$
View full question & answer→Question 543 Marks
Solve the following quadratic equations by factorization:
$x^2-x-a(a+1)=0$
Answer$x^2-x-a(a+1)=0$
$⇒ x^2+ {(a) - (a + 1)}x - a(a + 1) = 0$
${ 1 = a - a + 1}$
$⇒ x^2+ ax - (a + 1)x - a(a + 1) = 0$
$⇒ x(x + a) - (a + 1)(x + a) = 0$
$⇒ (x + a)(x - a - 1) = 0$
Either $x + a = 0,$ then $x = -a$
or $x - a - 1 = 0,$ then $x = a + 1$
$\therefore$ Roots are $-a, a + 1$
View full question & answer→Question 553 Marks
Solve the following quadratic equations by factorization:
$5 x^2-3 x-2=0$
AnswerWe have,
$5 x^2-3 x-2=0$
$\Rightarrow 5 x^2-5 x+2 x-2=0$
$\Rightarrow 5 x(x-1)+2(x-1)=0$
$\Rightarrow(x-1)(5 x+2)=0$
$\Rightarrow(x-1)=0 \text { or } 5 x+2=0$
$⇒ x = 1$ or $\text{x}=-\frac{2}{5}$
$\therefore$ x = 1 and $\text{x}=-\frac{2}{5}$ are the two roots of the given equation.
View full question & answer→Question 563 Marks
The product of Ramu's age $($in years$)$ five years ago and his age $($in years$)$ nine years later is $15$. Determine Ramu's present age.
AnswerLet the present age of Ramu be a $x$ years
Given that,
The product of his age years ago and his age $y$ nine years later is $15.$
Now, Ramu's age five years ago $= (x - 5)$ years
And Ramu's age nine years later $= (x + 9)$ years
Given that,
$(x-5)(x+9)=15$
$\Rightarrow x^2+9 x-5 x-45=15$
$\Rightarrow x^2+4 x-60=0$
$\Rightarrow x^2+10 x-6 x-60=0$
$\Rightarrow x(x+10)-6(x+10)=0$
$\Rightarrow(x+10)(x-6)=0$
$\Rightarrow x+10=0 \text { or } x-6=0$
$\Rightarrow x=-10 \text { or } x=6$
Since, age cannot be in negative value , so $x = 6$ years
$\therefore$ The present age of Ramu is $6$ years.
View full question & answer→Question 573 Marks
Find two consecutive odd positive integers, sum of whose squares is $970.$
AnswerLet two consecutive positive integers be $x$ and $x + 2$
$\text { A.T.Q. }(x)^2+(x+2)^2=970$
$\Rightarrow x^2+x^2+4 x+4-970=0$
$\Rightarrow 2 x^2+4 x-966=0$
$\Rightarrow x^2+2 x-483=0$
$\Rightarrow x^2+23 x-21 x-483=0$
$\Rightarrow x(x+23)-21(x+23)=0$
$\Rightarrow(x-21)(x+23)=0$
$\text { Either } x-21=0 \text { or } x+23=0$
$x = 21$ or $x = -23 ($rejected being -ve$)$
As integers should be +ve
$x = 21$ and $x + 2 = 21 + 2 = 23$
Hence integers are $21, 23$
View full question & answer→Question 583 Marks
Solve the following quadratic equations by factorization:
$\text{x}-\frac{1}{\text{x}}=3,\text{x}\neq0$
Answer$\text{x}-\frac{1}{\text{x}}=3$
$\Rightarrow\text{x}^2-3\text{x}-1=0$
Here $\text{a}=1,\text{b}=-3,\text{c}=-1$
$\therefore\text{D}=\text{b}^2-4\text{ac}$
$=(-3)^2-4\times(1)(-1)$
$=9+4=13$
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\text{x}={-(-3) \pm \sqrt{13} \over 2\times1}$
$\text{x}=\frac{3\pm\sqrt{13}}{2}$
View full question & answer→Question 593 Marks
The sum of the squares of two consecutive multiples of $7$ is $637.$ Find the multiples.
AnswerLet one of the number be $7x$ then the other number be $7(x + 1)$
Then according to question,
$\Rightarrow(7 x)^2+[7(x+1)]^2=637$
$\Rightarrow 49 x^2+49\left(x^2+2 x+1\right)=637$
$\Rightarrow 49 x^2+49 x^2+98 x+49-637=0$
$\Rightarrow 98 x^2+98 x-588=0$
$\Rightarrow x^2+x-6=0$
$\Rightarrow x^2+3 x-2 x-6=0$
$\Rightarrow x(x+3)-2(x+3)=0$
$\Rightarrow(x-2)(x+3)=0$
$\Rightarrow x-2=0 \text { or } x+3=0$
$\Rightarrow x=2 \text { or } x=-3$
Since, the numbers are mutiples of $7,$
Therefore, one number $= 7 × 2 = 14$
Then another number will be $7(x + 1) = 7 × 3 = 21$
Thus, the two consecutive multiples of $7$ are $14$ and $21$
View full question & answer→Question 603 Marks
The product of two successive integral multiples of $5$ is $300.$ Determine the multiples.
AnswerGiven that the product of two successive integral multiples of $5$ is $300$
Let the integers be $5x,$ and $5(x + 1)$
Then, ny the integers be $5x$ and $5(5x + 1)$
Then, by the hypothesis, we have
$5 x \times 5(5 x+1)=300$
$\Rightarrow 25 x(x+1)=300$
$\Rightarrow x^2+x=12$
$\Rightarrow x^2+x-12=0$
$\Rightarrow x^2+4 x-3 x-12=0$
$\Rightarrow x(x+4)-3(x+4)=0$
$\Rightarrow(x+4)(x-3)=0$
$\Rightarrow x=-4 \text { or } x=3$
$\text { If } x=-4,5 x=-20,5(x+1)=-15$
$x=3,5 x=15,5(x+1)=20$
The two successive integral multiples are $15, 20$ or $-15, -20$
View full question & answer→Question 613 Marks
Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
$3\text{x}^2+11\text{x}+10=10$
AnswerWe have been given that,
$3\text{x}^2+11\text{x}+10=0$
Now, divide throughout by $3$ we get,
$\text{x}^2+\frac{11}{3}\text{x}+\frac{10}{3}=0$
Now take the constant term to the $R.H.S.$ and we get,
$\text{x}^2+\frac{11}{3}\text{x}=-\frac{10}{3}$
Now add square of half co-efficient of $'x'$ on both the sides. we have,
$\text{x}^2+\frac{11}{3}\text{x}+\Big(\frac{11}{6}\Big)^2=\Big(\frac{11}{6}\Big)^2-\frac{10}{3}$
$\text{x}^2+\Big(\frac{11}{6}\Big)^2+2\Big(\frac{11}{3}\Big)\text{x}=\frac{1}{36}$
$\Big(\text{x}^2+\frac{11}{6}\Big)^2=\frac{1}{36}$
Since $R.H.S.$ is a positive number, therefore the roots of the equation exist.
So, now take the square root on both the sides and we get,
$\text{x}+\frac{11}{6}=\pm\frac{1}{6}$
$\text{x}=-\frac{11}{6}\pm\frac{1}{6}$
Now, we have the values of $'x'$ as
$\text{x}=-\frac{11}{6}\pm\frac{1}{6}$
$\text{x}=-\frac{5}{3}$
Also we have,
$\text{x}=-\frac{11}{6}-\frac{1}{6}$
$\text{x}=-2$
Therefore the roots of the equation are $-2$ and $-\frac{5}{3}$
View full question & answer→Question 623 Marks
In the following, determine whether the given values are solution of the given equation or not:
$x^2-3 x+2=0 . x=2, x=-1$
Answer$x^2-3 x+2=0 . x=2, x=-1$
Here, L.H.S. $=x^2-3 x+2$ and R.H.S. $=0$
Now, substitute $x=2$ in L.H.S.
We get $(2)^2-3(2)+2=4-6+2$
$ =6-6 $
$=0$
R.H.S.
Since, L.H.S. $=$ R.H.S.
$x=2$ is a solution for the given equation.
Similarly,
Now substitute $x=-1$ in L.H.S.
We get $(-1)^2-3(-1)+2$
$1+3+2=6 \neq \text { R.H.S. }$
Since L.H.S $\neq$ R.H.S.
$x=-1$ is not a solution for thr given equatoion.
View full question & answer→Question 633 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$2\text{x}^2+5\sqrt{3}\text{x}+6=0$
Answer$2\text{x}^2+5\sqrt{3}\text{x}+6=0$
Here, $\text{a}=2,\text{b}=5\sqrt{3},\text{c}=6$
$\therefore$ Discriminant $(\text{D})=\text{b}^2-4\text{ac}$
$=(5\sqrt{3})^2-4\times2\times6$
$=75-48=27$
$\because\text{D}>0$
$\therefore$ Roots are real and unequal
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-5\sqrt{3}\pm\sqrt{27}}{2\times2}$
$=\frac{-5\sqrt{3}\pm3\sqrt{3}}{4}$
$\therefore\text{x}=\frac{-5\sqrt{3}+3\sqrt{3}}{4}=\frac{-2\sqrt{3}}{4}$
$\text{x}=\frac{-\sqrt{3}}{2}$
and $\text{x}=\frac{-5\sqrt{3}-3\sqrt{3}}{4}=\frac{-8\sqrt{3}}{4}$
$\text{x}=-2\sqrt{3}$
Roots are $\frac{-\sqrt{3}}{2},-2\sqrt{3}$
View full question & answer→Question 643 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$2\text{x}^2-2\sqrt{2}\text{x}+1=0$
Answer$2\text{x}^2-2\sqrt{2}\text{x}+1=0$
$\Rightarrow(\sqrt{2}\text{x})^2-2\times\sqrt{2}\text{x}\times1+(1)^2=0$
$\Rightarrow(\sqrt{2}\text{x}-1)^2=0$
$\Rightarrow\sqrt{2}\text{x}-1=0$
$\Rightarrow\sqrt{2}\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{\sqrt{2}}$
$\therefore\text{x}=\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}$
View full question & answer→Question 653 Marks
Find the value of k for which root are real and equal in the following equations:
$2 k x^2-40 \mathrm{x}+25=0$
AnswerThe given quadric equation is $2 k x^2-40 \mathrm{x}+25=0$, and roots real and equal
Then find the value of k.
Here, $a = 2k, b = -40$ and $c = 25$
As we know that $D = b^2- 4ac$
Putting the value of $a = 2k, b = -40$ and $c = 25$
$= (-40)^2- 4 × 2k × 25$
$= 1600 - 200k$
The given equation will have real and equal roots , if $D = 0$
Thus,$ 1600 - 200k = 0$
$200k = 1600$
$\text{k}=\frac{1600}{200}$
$k = 8$
Therefore, the value of $k = 8$
View full question & answer→Question 663 Marks
John and javanti together have $45$ marbles. Both of them lost $5$ marbles each, and the product of the number of marber of marbles they now have is $128.$ From the quadratic equation to find how many marbles had to start with, if john and $x$ marbles.
AnswerIt is given that John had $'x'$ marbles.
We are also given that both John and javanti had $45$ marbles together.
So, Javanti should have $'45 - x'$ marbles with her.
Now, it is given taht both of them lose 5 marbles each.
So, in the new situation, John will have $' x - 5'$ marbles and Javanti will have $'45 - x - 5'$ marbles.
Also it is given taht the product of the number of marbles pof marbles both of them now is $128.$
Therefore,
$ (x-5)(45-x-5)=128 $
$ (x-5)(40-x)=128$
$ 40 x-x^2-200+5 x-128=0$
$ x^2-45 x+200+128=0 $
$ x^2-45 x+328=0$
Hence, this is the required quadratic equation.
View full question & answer→Question 673 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$3\text{x}^2+2\sqrt{5}\text{x}-5=0$
Answer$3\text{x}^2+2\sqrt{5}\text{x}-5=0$
Here $\text{a}=3,\text{b}=2\sqrt{5},\text{c}=-5$
$\therefore$ Discriminant $(\text{D})=\text{b}^2-4\text{ac}$
$=(2\sqrt{5})^2-4\times3\times(-5)$
$=20+60=80$
$\because\text{D > 0}$
$\therefore$ Roots are real and unequal
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-2\sqrt{5}\pm\sqrt{80}}{2\times3}$
$=\frac{-2\sqrt{5}\pm4\sqrt{5}}{6}$
$=\frac{-\sqrt{5}\pm2\sqrt{5}}{3}$
$\therefore\text{x}=\frac{-\sqrt{5}+2\sqrt{5}}{3}$
$\text{x}=\frac{\sqrt{5}}{3}$
and $\text{x}=\frac{-\sqrt{5}-2\sqrt{5}}{3}$
$=\frac{-3\sqrt{5}}{3}$
$\text{x}=-\sqrt{5}$
Roots are $-\sqrt{5},\frac{\sqrt{5}}{3}$
View full question & answer→Question 683 Marks
Three consecutive positive integers are such that the sum of the square of the first and the product of other two is $46,$ find the integers.
AnswerLet first number $= x$
Then second number $= x + 1$
and third number $= x + 2$
According co the condition,
$(x)^2+(x+1)(x+2)=46 $
$\Rightarrow x^2+x^2+3 x+2=46$
$\Rightarrow 2 x^2+3 x+2-46=0$
$\Rightarrow 2 x^2+3 x-44=0$
$\Rightarrow 2 x^2+11 x-8 x-44=0$
$\Rightarrow x(2 x+11)-4(2 x+11)=0$
$\Rightarrow(2 x+11)(x-4)=0$
Either $2x + 11 = 0$, then $\text{x}=\frac{-11}{2}$ which is not possible being fraction or $x - 4 = 0$, then $x = 4$
Numbers are $4, 5, 6$
View full question & answer→Question 693 Marks
The sum of two numbers is $18.$ The sum of their reciprocals is $\frac{1}{4}$ Find the numbers.
AnswerSum of two numbers $= 18$
Let one number $= x$
Then second number $= 18 - x$
According to the condition,
$\frac{1}{\text{x}}+\frac{1}{18-\text{x}}=\frac{1}{4}$
$\Rightarrow\frac{18-\text{x}+\text{x}}{\text{x}(18-\text{x})}=\frac{1}{4}$
$\Rightarrow\frac{18}{18\text{x}-\text{x}^2}=\frac{1}{4}$
$72=18 x-x^2$
$x^2-18 x+72=0$
$\Rightarrow x^2-12 x-6 x+72=0$
$\Rightarrow x(x-12)-6(x-12)=0$
$\Rightarrow(x-12)(x-6)=0$
Either $x - 12 = 0$, then $x = 12$
or $x - 6 = 0,$ then $x = 6$
- If $x = 12$, then
First number $= 12$ and second number $= 18 - 12 = 6$
- If $x = 6,$ then
First number $= 6$ then second number $= 18 - 6 = 12$
Number are $6, 12$ View full question & answer→Question 703 Marks
Solve the following quadratic equations by factorization:
$a b x^2+\left(b^2-a c\right) x-b c=0$
AnswerWe have,
$a b x^2+\left(b^2-a c\right) x-b c=0$
${\left[a b x-b c=-a b^2 c \Rightarrow-a b^2 c=b^2 \times-a c \text { and } b^2-a c=b^2+(-a c)\right]}$
$\Rightarrow a b x^2+b^2 x-a c x-b c=0$
$\Rightarrow b x(a x+b)-c(a x+b)=0$
$\Rightarrow(a x+b)(b x-c)=0$
$\Rightarrow a x+b=0 \text { or } b x-c=0$
$\Rightarrow\text{x}=-\frac{\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{c}}{\text{b}}$
$\therefore\text{x}=-\frac{\text{b}}{\text{a}}$ and $\text{x}=\frac{\text{c}}{\text{b}}$ are the two roots the given equation.
View full question & answer→Question 713 Marks
Write the value of k for which the quadratic equation $x^2-k x+4=0$ has equal roots.
AnswerThe given quadric equation is $x^2-k x+4=0$, and roots are equal.
Then find the value of $k$.
Here, $\mathrm{a}=1, \mathrm{~b}=-\mathrm{k}$ and $\mathrm{c}=4$
As we know that $D=b^2-4 a c$
Putting the value of $\mathrm{a}=1, \mathrm{~b}=-\mathrm{k}$ and $\mathrm{c}=4$
$ =(-k)^2-4 \times 1 \times 4 $
$ =k 2-16$
The given equation will have equal roots, if $D=0$
$k^2-16=0 $
$ k^2=16$
$\text{k}=\sqrt{16}$
$\text{k}=\pm4$
Therefore, the value of $\text{k}=\pm4$
View full question & answer→Question 723 Marks
The difference of squares of two numbers is $180.$ The square of the smaller number is 8 times the larger number. Find two numbers.
AnswerLet the number be $x$
By the given hypothesis, we have
$\text { According to question } \mathrm{x}^2-\mathrm{y}^2=180 \text { and } \mathrm{y}^2=8 \mathrm{x}$
$\Rightarrow x^2-8 x=180$
$\Rightarrow x^2-8 x-180=0$
$\Rightarrow x^2+10 x-18 x-180=0$
$\Rightarrow x(x+10)-18(x+10)=0$
$\Rightarrow(x+10)(x-18)=0$
$\Rightarrow x=-10 \text { or } x=18$
Case I: $x = 18$
$⇒ 8x = 8 × 18 = 144$
Lerger number $=\sqrt{144}=\pm12$
Case II: $x = -10$
Square of larger number $8x = -80$ here no perfect square exist, hence the numbers are $18, 12.$
View full question & answer→Question 733 Marks
Solve the following quadratic equations by factorization:
$25x(x + 1) = -4$
AnswerWe have been given
$25 x(x+1)=-4$
$25 x^2+25 x+4=0$
$25 x^2+20 x+5 x+4=0$
$5 x(5 x+4)+1(5 x+4)=0$
$(5 x+1)(5 x+4)=0$
Therefore,
$5x + 1 = 0$
$5x = -1$
$\text{x}=\frac{-1}{5}$
Or,$ 5x + 4 = 0$
$5x = -4$
$\text{x}=\frac{-4}{5}$
Hence, $\text{x}=\frac{-1}{5}$ or $\text{x}=\frac{-4}{5}$
View full question & answer→Question 743 Marks
Solve the following quadratic equations by factorization:
$ 6 x^2+11 x+3=0 $
AnswerWe have been given
$ 6 x^2+11 x+3=0 $
$ 6 x^2+9 x+2 x+3=0 $
$ 3 x(2 x+3)+1(2 x+3)=0 $
$ (2 x+3)(3 x+1)=0 $
$ 2 x+3=0$
$\text{x}=\frac{-3}{2}$
Or, $3x + 1 = 0$
$\text{x}=\frac{-1}{3}$
Hence, $\text{x}=\frac{-3}{2}$ or $\text{x}=\frac{-1}{3}$
View full question & answer→Question 753 Marks
A two-digit number is such that the product of its digits is $8.$ When $18$ is subtracted from the number, the digits interchange their places. Find the number.
AnswerLet the two digits be $x$ and $x - 2$
Given that the product of their digit is $8$
$ \Rightarrow x(x-2)=8$
$\Rightarrow x^2-2 x-8=0$
$\Rightarrow x^2-4 x+2 x-8=0$
$\Rightarrow x(x-4)+2(x-4)=0$
$\Rightarrow(x-4)(x+2)=0$
$\Rightarrow x=4 \text { or } x=-2$
Considering the positive value $x = 4, x - 2 = 2$
$\therefore$ The two digit number is $42$
View full question & answer→Question 763 Marks
Write the sum of real roots of the equation $x^2+ |x| - 6 = 0$
AnswerThe given quadric equation is $x^2+ |x| - 6 = 0$
Here, $\text{a}=1,\text{b}=\pm1$ and $\text{c}=-6$
As we know that $\text{D}=\text{b}^2-4\text{ac}$
Putting the value of $\text{a}=1,\text{b}=\pm1$ and $\text{c}=-6$
$=(\pm1)^2-4\times1\times-6$
$=1+24$
$=25$
Since, $\text{D}\geq0$
Therefore, root of the given equation are real and distinct.
Thus, sum of the roots be $= 0$
View full question & answer→Question 773 Marks
Solve the following quadratic equations by factorization:
$\text{x}^2-\Big(\sqrt{3}+1\Big)\text{x}+\sqrt{3}=0$
Answer$\text{x}^2-\Big(\sqrt{3}+1\Big)\text{x}+\sqrt{3}=0$
$\Rightarrow\text{x}^2-\sqrt{3}\text{x}-\text{x}+\sqrt{3}=0$
$\Rightarrow\text{x}\Big(\text{x}-\sqrt{3}\Big)-1\Big(\text{x}-\sqrt{3}\Big)=0$
$\Rightarrow\Big(\text{x}-\sqrt{3}\Big)(\text{x}-1)=0$
Either $\text{x}-\sqrt{3}=0,$ then $\text{x}=\sqrt{3}$
or $\text{x}-1=0,$ then $\text{x}=1$
$\therefore$ Roots are $\text{x}=\sqrt{3},1$
View full question & answer→Question 783 Marks
Determine the nature of the root of following quadratic equation:
$9 a^2 b^2 x^2-24 a b c d x+16 c^2 d^2=0$, $\text{a}\neq0,\text{b}\neq0$
AnswerThe given quadric equation is $9 a^2 b^2 x^2-24 a b c d x+16 c^2 d^2=0$
Here, $a=9 a^2 b^2, b=-24 a b c d$ and $c=16 c^2 d^2$
As we know that $D=b^2-4 a c$
Putting the value of $a=9 a^2 b^2, b=-24 a b c d$ and $c=16 c^2 d^2$
$ =(24 a b c d)^2-4 \times 9 a^2 b^2 \times 16 c^2 d^2 $
$ =\left(576 a^2 b^2 c^2 d^2\right)-576 a^2 b^2 c^2 d^2$
$= 0$
Since, $D = 0$
Therefore, root of the given equation are real and equal.
View full question & answer→Question 793 Marks
Solve the following quadratic equations by factorization:
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3},$ $\text{x}\neq2, 4$
AnswerWe have been given
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3}$
$3\left(x^2-5 x+4+x^2-5 x+6\right)=10\left(x^2-6 x+8\right) $
$4 x^2-30 x+50=0 $
$2 x^2-15 x+25=0 $
$2 x^2-10 x-5 x+25=0 $
$2 x(x-5)-5(x-5)=0 $
$(2 x-5)(x-5)=0$
Therefore,
$2x - 5 = 0$
$2x = 5$
$\text{x}=\frac{5}{2}$
or, $x - 5 = 0$
$x = 5$
Hence, $\text{x}=\frac{5}{2}$ or $x = 5$
View full question & answer→Question 803 Marks
The sum of squares of two consecutive odd positive integers is $394.$ Find them.
AnswerLet two consecutive odd positive integer be $(2x - 1)$ and other $(2x + 1)$
Then according to question,
$ (2 x+1)^2+(2 x-1)^2=394 $
$ 4 x^2+4 x+1+4 x^2-4 x+1=394 $
$ 8 x^2+2=394 $
$ 8 x^2=394-2$
$\text{x}^2=\frac{392}{8}$
$\text{x}^2=49$
$\text{x}=\sqrt{49}$
$\text{x}=\pm7$
Since, $x$ beging a positive numner, so $x$ cannot be negative.
Therefore,
When x = 7 then odd positive
$2x - 1 = 2 × 7 - 1$
$2x - 1 = 13$
And, $2x + 1 = 2 × 7 + 1$
$2x + 1 = 15$
Thus, two consecutive odd positive integer be $ 13, 15$
View full question & answer→Question 813 Marks
Since, $\text{x}=-\frac{1}{2},$ is a solution of the quadratic equation $3x^2+ 2kx - 3 = 0$, find the value of k.
AnswerSince, $\text{x}=-\frac{1}{2},$ is a solution of the quadratic equation $3x^2+ 2kx - 3 = 0$
So, it satisfies the given equation,
$\therefore3\Big(-\frac{1}{2}\Big)^2+2\text{k}\Big(-\frac{1}{2}\Big)-3=0$
$\Rightarrow\frac{3}{4}-\text{k}-3=0$
$\Rightarrow\text{k}=\frac{3}{4}-3$
$\Rightarrow\text{k}=\frac{3-12}{4}$
$\Rightarrow\text{k}=-\frac{9}{4}$
Thus, the value k is $-\frac{9}{4}$
View full question & answer→Question 823 Marks
Solve the following quadratic equations by factorization:
$x^2+2 a b=(2 a+b) x$
AnswerWe have
$x^2+2 a b=(2 a+b) x$
$\Rightarrow x^2-(2 a+b) x+2 a b=0$
${[\because 2 a b=-8 a x-b \Rightarrow-(8 a+b)=-8 a-b}$
$\Rightarrow x^2-2 a x-b x+2 a b=0$
$\Rightarrow x-(x-8 a)-b(x-2 a)=0$
$\Rightarrow(x-8 a)(x-b)=0$
$\Rightarrow x-8 a=0 \text { or } x-b=0$
$\Rightarrow x=8 a=0 \text { or } x=b$
$\therefore x = 8a$ and $x = b$ are the two roots of the given equation.
View full question & answer→Question 833 Marks
Solve the following quadratic equations by factorization:
$3 x^2=-11 x-10$
Answer$3 x^2=-11 x-10$
$ \Rightarrow 3 \mathrm{x}^2+11 \mathrm{x}+10=0$
$ \Rightarrow 3 \mathrm{x}^2+11 \mathrm{x}+10=0$
$\begin{cases}\because3\times10=30\\\therefore30=5\times6\\11=5+6\end{cases}$
$\Rightarrow 3 x^2+6 x+5 x+10=0$
$⇒ 3x(x + 2) + 5(x + 2) = 0$
$⇒ (x + 2)(3x + 5) = 0$
Either $x + 2 = 0$, then $x = -2$
Or $3x + 5 = 0$. then $3x = -5$
$\Rightarrow\text{x}=\frac{-5}{3}$
$\therefore$ Roots are $x = -2, \frac{-5}{3}$
View full question & answer→Question 843 Marks
In the following, determine the set of values of k for which the given quadratic equation has real root:
$2 x^2+3 x+k=0$
AnswerThe given equation is $2 x^2+3 x+k=0$
Given that the quadratic equation has real roots i.e.,
$\text{D}=\text{b}^2-4\text{ac}\geq0$
Given here $\text{a}=2,\text{b}=3,\text{c}=\text{k}$
$\Rightarrow9-4\times2\times\text{k}\geq0$
$\Rightarrow9-8\text{k}\geq0$
$\Rightarrow9\geq8\text{k}$
$\Rightarrow\text{k}\leq\frac{9}{8}$
The value of k does not exceed $\frac{4}{8}$ to have roots.
View full question & answer→Question 853 Marks
Find the value of k for which root are real and equal in the following equations:
$4 x^2-3 k x+1=0$
Answer$4 x^2-3 k x+1=0$
Here $a=4, b=-3 k, c=1$
$ \therefore \text { Discriminant }(D)=b^2-4 a c$
$=(-3 k)^2-4 \times 4 \times 1$
$=9 k^2-6$
$\therefore$ Roots are real and equal
$ \therefore D=0 $
$ \Rightarrow 9 k^2-16=0 $
$ \Rightarrow 9 k^2=16$
$\Rightarrow\text{k}^2=\frac{16}{9}$
$\text{k}=\Big(\pm\frac{4}{3}\Big)^2$
$\therefore\text{k}=\pm\frac{4}{3}$
View full question & answer→Question 863 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$ 16 x^2=24 x+1$
Answer$ 16 x^2=24 x+1$
$\Rightarrow 16 x^2-24 x-1=0$
$\therefore \text { Discriminate }=b^2-4 a c$
$=(-24)^2-4 \times 16 \times(-1)$
$=576+64=640$
$\because D>0$
$\therefore$ Roots are real and unequal
$\therefore\text{x} = {-\text{b}\pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-24)\pm\sqrt{640}}{2\times16}$
$=\frac{24\pm8\sqrt{10}}{32} ($Dividng by $8)$
$=\frac{3\pm\sqrt{10}}{4}$
$\therefore$ Roots are $\frac{3+\sqrt{10}}{4},\frac{3-\sqrt{10}}{4}$
View full question & answer→Question 873 Marks
Solve for x.
$\frac{1}{\text{x}}+\frac{2}{2\text{x}-3}=\frac{1}{\text{x}-2},$ $\text{x}\neq0,\frac{3}{2},2$
Answer$\frac{1}{\text{x}}+\frac{2}{2\text{x}-3}=\frac{1}{\text{x}-2},$ $\text{x}\neq0,\frac{3}{2},2$
$\Rightarrow\frac{(2\text{x}-3)+2\text{x}}{\text{x}(2\text{x}-3)}=\frac{1}{(\text{x}-2)}$
$ \Rightarrow(x-2)(4 x-3)=x(2 x-3)$
$\Rightarrow 4 x^2-11 x+6=2 x^2-3 x$
$\Rightarrow 2 x^2-8 x+6=0$
$\Rightarrow x^2-4 x+3=0$
$\Rightarrow x^2-3 x-x+3=0$
$\Rightarrow x(x-3)-(x-3)=0$
$\Rightarrow(x-1)(x-3)=0$
$\therefore x-1=0$
$\Rightarrow x=1$
$\text { and } x-3=0$
$\Rightarrow x=3$
View full question & answer→Question 883 Marks
A passenger train takes one hour less for a journey of $150\ km$ if its speed is increased by $5\ km/hr$ from its usual speed. Find the usual speed of the train.
AnswerLet the usual speed of train be $x\ km/hr$ then
Increased speed of the train $= (x +5)\ km/hr$
Time taken by the train under usual speed to cover $150\ km =\frac{150}{\text{x}}\text{hr}$
Time taken by the train under increased speed to cove $150\ km =\frac{150}{\text{(x+5)}}\text{hr}$
Therefore,
$\frac{150}{\text{x}}-\frac{150}{(\text{x}+5)}=1$
$\frac{\{150(\text{x+5})-150\text{x}\}}{\text{x}(\text{x}+5)}=1$
$\frac{150\text{x}+750-150\text{x}}{\text{x}^2+5\text{x}}=1$
$750=x^2+5 x$
$x^2+5 x-750=0$
$x^2-25 x+30 x-750=0$
$x(x-25)+30(x-25)=0$
$(x-25)(x+30)=0$
$\text { So, either }$
$(x-25)=0$
$x=25$
$\text { Or }(x+30)=0$
$x=-30$
But, the speed of the the train can never be negative,
Hence, the usual speed of train is $x = 25\ km/hr$
View full question & answer→Question 893 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$2\text{x}^2-2\sqrt{6}\text{x}+3=0$
Answer$2\text{x}^2-2\sqrt{6}\text{x}+3=0$
Here $\text{a}=2,\text{b}=-2\sqrt{6}$ and $\text{c}=3$
$\therefore$ Discriminant $(\text{D})=\text{b}^2-4\text{ac}$
$=(-2\sqrt{6})^2-4\times2\times3$
$=24-24=0$
$\because\text{D}=0$
$\therefore$ Roots are real and equal
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-2\sqrt{6})\pm0}{2\times2}$
$=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}$ or $\frac{\sqrt{6}}{\sqrt{2}\times\sqrt{2}}=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{2}\times\sqrt{2}}$
$=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}$
$\therefore$ Roots are $=\frac{\sqrt{3}}{\sqrt{2}},\sqrt{\frac{3}{2}}$
View full question & answer→Question 903 Marks
Determine the set of values of k for which the given following quadratic has real root:
$4 x^2-3 k x+1=0$
AnswerThe given quadric equation is $4 x^2-3 k x+1=0$, and roots are real.
Then find the value of k.
Here, $a= 4, b = -3k$ and $c = 1$
As we know that $D = b^2- 4ac$
Putting the value of $a = 4, b = -3k$ and $c = 1$
$=(-3 k)^2-4 \times 4 \times 1$
$=9 k^2-16$
The given equation will have real roots, if $\text{D}\geq0$
$9\text{k}^2-16\geq0$
$9\text{k}^2\geq16$
$\text{k}^2\geq\frac{16}{9}$
$\text{k}\geq\sqrt{\frac{16}{9}}$
$\text{k}\leq-\frac{4}{3}$ or $\text{k}\geq\frac{4}{3}$
Therefore, the value of $\text{k}\leq-\frac{4}{3}$ or $\text{k}\geq\frac{4}{3}$
View full question & answer→Question 913 Marks
A two digit number is $4$ times the sum of its digits and twice the product of its digits. Find the number.
AnswerLet the require digit be $= (10x + y)$
Then according to question,
$(10x + y) = 4(x + y)$
$(10x + y) = 4x + 4y$
$10x + y - 4x - 4y = 0$
$6x - 3y = 0$
$2x - y = 0$
$2x = y ....(i)$
And,$ (10x + y) = 2xy ....(ii)$
Now, putting the value of $y$ in equation $(ii)$ from $(i)$
$(10x + 2x) = 2x × 2x$
$4x^2- 12x = 0$
$4x(x - 3) = 0$
$x(x - 3) = 0$
So, either
$x = 0$
Or $(x - 3) = 0$
$x = 3$
So, the digite can never be negative.
When $x = 3$ then
$y = 2x$
$y = 2 × 3$
$y = 6$
Therefore, number
$= 10x + y$
$= 10 × 3 + 6$
$= 36$
Thus, the required number be $36$
View full question & answer→Question 923 Marks
Solve the following quadratic equations by factorization:
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
AnswerWe have been given
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}^2-4\text{x}+\text{x}-2\sqrt{2}=0$
$\sqrt{2}\text{x}\big(\text{x}-2\sqrt{2}\big)+1\big(\text{x}-2\sqrt{2}\big)=0$
$\big(\text{x}-2\sqrt{2}\big)\big(\sqrt{2}\text{x}+1\big)=0$
Therefore,
$\text{x}-2\sqrt{2}=0$
$\text{x}=2\sqrt{2}$
or, $\sqrt{2}\text{x}+1=0$
$\sqrt{2}\text{x}=-1$
$\text{x}=\frac{-1}{\sqrt{2}}$
Hence, $\text{x}=2\sqrt{2}$ or $\text{x}=\frac{-1}{\sqrt{2}}$
View full question & answer→Question 933 Marks
The difference of squares of two number is $88.$ If the larger number is $5$ less than twice the smaller number, then find the two numbers.
AnswerLet teh smaller numbers be $x$
Then according to question,
The larger number be $= 2x - 5$, then
$ (2 x-5)^2-x^2=88$
$4 x^2-20 x+25-x^2-88=0$
$3 x^2-20 x-63=0$
$3 x^2-27 x+7 x-63=0$
$3 x(x-9)+7(x-9)=0$
$(x-9)(3 x+7)=0$
$(x-9)=0$
$x=9$
$\text { Or }(3 x+7)=0$
$\text{x}=\frac{-7}{3}$
Since,$x$ being a positive integer so, $x$ cannot be negative,
Therefore,
When $x = 9$ then larger number be
$2x - 5 = 2 × 9 - 5$
$2x - 5 = 18 - 5$
$2x - 5 = 13$
Thus, two consecutive number be either $9, 13$
View full question & answer→Question 943 Marks
Find the whole number which when decreased by $20$ is equal to $69$ times the reciprocal of the number.
AnswerLet the whole numbers be $x.$
Then according to question,
$(\text{x}-20)=69\times\frac{1}{\text{x}}$ $x(x-20)=69$
$x^2-20 x-69=0$
$x^2-23 x+3 x-69=0$
$x(x-23)+3(x-23)=0$
$(x-23)(x+3)=0$
$(x-23)=0$
$x=23$
$\text { Or }(x+3)=0$
$x=-3$
Since, whole numbers being a positive, so $x$ cannot be negative.
Thus, whole numbers be $23$
View full question & answer→Question 953 Marks
Solve the following quadratic equations by factorization:
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
AnswerWe have been given
$3\text{x}^2-2\sqrt{6}\text{x}+2=0$
$3\text{x}^2-\sqrt{6}\text{x}-\sqrt{6}\text{x}+2=0$
$\sqrt{3}\text{x}\big(\sqrt{3}\text{x}-\sqrt{2}\big)-\sqrt{2}\big(\sqrt{3}\text{x}-\sqrt{2}\big)=0$
$\big(\sqrt{3}\text{x}-\sqrt{2}\big)(\sqrt{3}\text{x}-\sqrt{2})=0$
Therefore,
$\sqrt{3}\text{x}-\sqrt{2}=0$
$\sqrt{3}\text{x}=\sqrt{2}$
$\text{x}=\sqrt{\frac{2}{3}}$
Hence, $\text{x}=\sqrt{\frac{2}{3}}$
View full question & answer→Question 963 Marks
If p, q are real and $\text{p}\neq\text{q},$ then show that the show that the roots of the equation $(p-q) x^2+5(p+q) x-2(p-q)=0$ are real and unequal.
AnswerThe quadric equation is $(p-q) x^2+5(p+q) x-2(p-q)=0$
Here, $a=(p-q), b=5(p+q)$ and $c=-2(p-q)$
As we know that $D=b^2-4 a c$
Putting the value of $a=(p-q), b=5(p+q)$ and $c=-2(p-q)$
$\Rightarrow D=\{5(p+q)\}^2-4(p-q)(-2(p-q))$
$\Rightarrow D=25\left(p^2+2 p q+q^2\right)+8\left(p^2-2 p q+q^2\right)$
$\Rightarrow D=25 p^2+50 p q+25 q^2+8 p^2-16 p q+8 q^2$
$\Rightarrow D=33 p^2+34 p q+33 q^2$
Since p and q are real and $\text{p}\neq\text{q},$ therefore, the value of $\text{D}\geq0$
Thus, the roots of the given equation are and unequal.
Hence, proved.
View full question & answer→Question 973 Marks
Solve for x.
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6},$ $\text{x}\neq3,-5$
Answer$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6},$ $\text{x}\neq3,-5$
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6}$
$\Rightarrow\frac{\text{x}+5-\text{x}+3}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow\frac{8}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow 48=x^2+2 x-15$
$\Rightarrow x^2+2 x-15-48=0$
$\Rightarrow x^2+2 x-63=0$
$\Rightarrow x^2+9 x-7 x-63=0$
$\Rightarrow x(x+9)-7(x+9)=0$
$\Rightarrow(x-7)(x+9)=0$
$\Rightarrow x=7,-9$
View full question & answer→Question 983 Marks
The sum of a number and its positive square is $\frac{63}{4},$ find the number.
AnswerLet first numbers be x
Then according to question
$\text{x}+\text{x}^2=\frac{63}{4}$
$ 4\left(x+x^2\right)=63$
$4 x^2+4 x-63=0$
$4 x^2+18 x-14 x-63=0$
$2 x(2 x+9)-7(2 x+9)=0$
$(2 x+9)(2 x-7)=0$
$(2 x+9)=0$
$\text{x}=-\frac{9}{2}$
Or $(2\text{x}-7)=0$
$\text{x}=\frac{7}{2}$
Thus, the required number be $\frac{7}{2},\frac{-9}{2}$
View full question & answer→Question 993 Marks
The product of two consecutive positive integers is $306.$ Form the quadratic equation to find the integers, if $x$ denotes the smaller integer.
AnswerGiven that the smallest integer of $2$ consecutive integer is denoted by $x.$
$⇒$ The two integer will be x and $(x + 1)$
Product of two integers $⇒ x(x + 1)$
Given that the product is $306$
$\therefore x(x + 1) = 306$
$ \Rightarrow x^2+x=306 $
$\Rightarrow x^2+x-306=0$
$\therefore$ The required quadratic equation is $x^2+ x - 306 = 0$
View full question & answer→Question 1003 Marks
Show that $x = -2$ is a solution of $3x^2+ 13x + 14 = 0.$
Answer$\text { Given that the equation of } 3 x^2+13 x+14=0$
$3 x^2+7 x+6 x+14=0$
$x(3 x+7)+2(3 x+7)=0$
$(3 x+7)(x+2)=0$
$(3 x+7)=0$
$\text{x}=\frac{-7}{3}$
Or $(x + 2) = 0$
$x = -2$
Therefore, $x = 2$ is the solution of given equation.
Hence, proved.
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