Question 13 Marks
Find the values of k for which the given quadratic equation has real and distinct root:
$kx^2 + 2x + 1 = 0$
Answer$kx^2 + 2x + 1 = 0$
Here, $a = k, b = 2, c = 1$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (2)^2 - 4 \times k \times 1$
$= 4 - 4k$
$\because$ Roots are real and distinct,
$\therefore D > 0$
$\Rightarrow 4 - 4k > 0$
$\Rightarrow 1 - k > 0$
$\Rightarrow 1 > k$
$\Rightarrow k < 1$
$\therefore$ k < 1
View full question & answer→Question 23 Marks
Find the value of k for which the following equation have real root:
$4x^2 + kx + 3 = 0$
AnswerThe given quadration is $4x^2 + kx + 3 = 0$, and roots are real and equal.Then find the value of p.
Here, $4x^2 + kx + 3 = 0$
$So, a = 4, b = p$ and $c = 3$
As know that $D = b^2 - 4ac$
$$Putting the value of $a = 4, b = p$ and $c = 3$
$D = (p)^2 - 4(4)(3)$
$D = p^2 - 48$
The given equation will have real and equal roots, if $D = 0$
So, $p2 - 48 = 0$
Now factorizing the above equation,
$p2 - 48 = 0$
$\Rightarrow\text{p}^2-(4\sqrt{3})^2=0$
$\Rightarrow(\text{p}-4\sqrt{3})(\text{p}+4\sqrt{3})=0$
$\Rightarrow\text{p}-4\sqrt{3}=0$ or $\text{p}+4\sqrt{3}=0$
$\Rightarrow\text{p}=4\sqrt{3}$ or $\text{p}=-4\sqrt{3}$
Therefore, the value of $\text{p}=\pm4\sqrt{3}$
View full question & answer→Question 33 Marks
Find the least positive value of k for which the equation $x^2 + kx + 4 = 0$ has real roots.
AnswerThe given equation is $x^2 + kx + 4 = 0$
Given that the equation has real roots
i.e., $\text{D}=\text{b}^2-4\text{ac}\geq0$
$\Rightarrow\text{k}^2-4\times1\times4\geq0$
$\Rightarrow\text{k}^2-16\geq0$
$\Rightarrow\text{k}\geq4$ or $\text{k}\leq-4$
The least positive value of k = 4, for the equation to have real roots.
View full question & answer→Question 43 Marks
Find the value of k for which the root are real and equal in the following equations:
$4x^2 + kx + 9 = 0$
AnswerThe given equation is $4x^2 + kx + 9 = 0$
This equation is in the form of $ax^2 + bx + c = 0$
Here, $a = 4, b = k$ and $c = 9$
Given that, the equation has real and equal roots
i.e., $D = b^2 - 4ac = 0$
$\Rightarrow k^2 - 4 \times 4 \times 9 = 0$
$\Rightarrow k^2 - 16 \times 9$
$\Rightarrow\text{k}=\sqrt{16\times9}$
$\Rightarrow k = 4 \times 3$
$\Rightarrow k = 12$
$\therefore$ The value of $k = 12$
View full question & answer→Question 53 Marks
Find the value of k for which the following equation have real root:
$kx(x - 3) + 9 = 0$
Answer$kx(x - 3) + 9 = 0$
$kx^2 - 3kx + 9 = 0$
Here, $a = k, b = -3k , c = 9$
$\therefore D = b^2 - 4ac$
$\Rightarrow D = (-3k)^2 - 4(k)9$
$\Rightarrow D = 9k^2 - 36k$
For roots to be real
$\Rightarrow D = 0$
$\Rightarrow 9k^2 - 36k = 0$
$\Rightarrow 9k(k - 4) = 0$
$\Rightarrow k - 4 = 0$
$\Rightarrow k = 4$
$\therefore k = 4$
View full question & answer→Question 63 Marks
In the following, determine whether the given values are solution of the given equation or not:
$x^2 + x + 1 = 0, x = 0, x = 1$
AnswerWe have been given that,
$x^2 + x + 1 = 0, x = 0, x = 1$
Now, if x = 0 is a solution of the equation then it should satisfy the equation.
So, substituting x = 0 in the equation we get
$x^2 + x + 1$
$= (0)^2 + (0) + 1$
$= 1$
Hence x = 0 is not a solution of the given quadratic equation.
Also, if x = 1 is a solution of the equation it should satisfy the equation So, substituting x = 1 in the equation, we get
$x^2 + x + 1$
$= (1)^2 + (1) + 1$
$= 3$
Hence x = 1 is not a soluion of the quadratic equation.
Therefore, from the above results we find out that both x = 0 and x = 1 are not a solution of the given quadratic equation.
View full question & answer→Question 73 Marks
Solve the following quadratic equations by factorization:$(a + b)^2 x^2 - 4abx - (a - b)^2 = 0$
AnswerWe have been given
$(a + b)^2 x^2 - 4abx - (a - b)^2 = 0$
$(a + b)^2 x^2 - (a + b)^2 x + - (a - b)^2 = 0$
$(a + b)^2 x(x - 1) + (a - b)^2 (x - 1) = 0$
$((a + b)^2 x + (a + b)^2) (x - 1) = 0$
Therefore,
$(a + b)^2 x + (a - b)^2 = 0$
$(a + b)^2 x = -(a - b)^2$^
$\text{x}=\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)^2$
or, $x - 1 = 0$
$x = 1$
Hence, $\text{x}=\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)^2$ or $x = 1$
View full question & answer→Question 83 Marks
If an integer is added to its square, the sum is $90$. Find the integer with the help of quadratic equation.
AnswerLet the integer be 'x'
Given that if an integer is added to its square, the sum is $90$
$\Rightarrow x + x^2 = 90$
$\Rightarrow x + x^2 - 90 = 0$
$\Rightarrow x^2 + 10x - 9x - 90 = 0$
$\Rightarrow x(x + 10) - 9(x + 10) = 0$
$\Rightarrow x = -10 or x = 9$
$\therefore$ The value of an integer are $-10$ or $9$
View full question & answer→Question 93 Marks
An express train takes 1 hour less than a passenger train to travel 132km between Mysore and Banglore. If the average speed of the express train is 11km/ hr more than that of the passenger train, from the quadratic equation to find the average speed of express train.
AnswerNow let us assume taht the speed of the speed of the express train be 'x' km/ hr. Therefore according to the question spreed of the passenger train will be 'x - 11'km/ hr. Now we know that the total distance travelled by both rthe trains was 132km.
We also know that so, the time by express train would be $\Big(\frac{132}{\text{x}}\Big)$ hr and the time taken by the passenger train would be $\Big(\frac{132}{\text{x}-11}\Big)$ he. Now, we also know that the express train took 1 hr less than the passenger.
Therefore, we have
$\Big(\frac{132}{\text{x}}\Big)=\Big(\frac{132}{\text{x}-11}\Big)-1$
$\Big(\frac{132}{\text{x}-11}\Big)-\Big(\frac{132}{\text{x}}\Big)=1$
$\Big(\frac{132\text{x}-132(\text{x}-11)}{\text{x}^2-11\text{x}}\Big)=1$
$\text{x}^2-11\text{x}-1452=0$
Therefore, this is the required equation.
View full question & answer→Question 103 Marks
Determine. if 3 is a root of the equation given below:
$\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}=\sqrt{4\text{x}^2-14\text{x}+16}$
Answer$\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}=\sqrt{4\text{x}^2-14\text{x}+16}$
If x = 3, then
L.H.S. $\sqrt{\text{x}^2-4\text{x}+3}+\sqrt{\text{x}^2-9}$
$=\sqrt{(3)^2-4\times3+3}+\sqrt{(3)^2-9}$
$=\sqrt{9-12+3}+\sqrt{9-9}$
$=\sqrt{0}+\sqrt{0}$
$=0$
R.H.S. $=\sqrt{4\text{x}^2-14\text{x}+16}$
$=\sqrt{4(3)^2-14\times3+16}$
$=\sqrt{4\times9-42+16}$
$=\sqrt{36+16-42}$
$=\sqrt{52-42}$
$=\sqrt{10}$
$\because$ L.H.S. $\neq$ R.H.S.
View full question & answer→Question 113 Marks
Write the value of $\lambda,$ for which $\text{x}^2+4\text{x}+\lambda$ is a perfect square.
AnswerIn $\text{x}^2+4\text{x}+\lambda$
$\text{a}=1,\text{b}=4,\text{c}=\lambda$
$\text{x}^2+4\text{x}+\lambda$ will be a perfect square if $\text{x}^2+4\text{x}+\lambda=0$ has equal roots
$\Rightarrow\text{D}=\text{b}^2-4\text{ac}$
$\Rightarrow\text{D}=(4)^2-4\times1\times\lambda$
$\Rightarrow\text{D}=0$
$\Rightarrow16-4\lambda=0$
$\Rightarrow16=4\lambda$
$\Rightarrow\lambda=4$
Hence $\lambda=4$
View full question & answer→Question 123 Marks
Solve the following quadratic by factorization:$a(x^2 + 1) - x(a^2 + 1) = 0$
AnswerWe have
$a(x^2 + 1) - x(a^2 + 1) = 0$
$\Rightarrow a(x^2 - 1) - a^2x - x + a = 0$
[ $\because$
$a \times a = a^2 $
$\Rightarrow a^2 = -a^2 \times -1-(a^2 + 1) = a^2 - 1]$
$\Rightarrow a \times (x - a) - 1(x - a) = 0$
$\Rightarrow (x - a)(ax - 1) = 0$
$\Rightarrow x - a = 0 $or ax $- 1 = 0$
$\Rightarrow x = a$ or $\text{x}=\frac{1}{\text{a}}$
$\therefore$ x = a and $\text{x}=\frac{1}{\text{a}}$ are the two roots of the given equation.
View full question & answer→Question 133 Marks
The difference of two natural numbers is $3$ and the difference of their reciprocals is $\frac{3}{28}.$ Find the numbers.
AnswerLet the larger tatural number be x
and the smaller natural number be y
$\Rightarrow x - y = 3 .....(i)$
$\Rightarrow x = 3 + y ......(ii)$
$\Rightarrow\frac{1}{\text{y}}-\frac{1}{\text{x}}=\frac{3}{28}$
$\Big[$ If x < y then $\frac{1}{\text{x}}<\frac{1}{\text{y}}\Big]$
$\Rightarrow\frac{\text{x}-\text{y}}{\text{xy}}=\frac{3}{28}$
$\Rightarrow\frac{3}{\text{xy}}=\frac{3}{28}$ [From (i)]
$\Rightarrow\frac{1}{\text{xy}}=\frac{1}{28}$
$\Rightarrow xy = 28$
$\Rightarrow (3 + y)y = 28 [From (ii)]$
$\Rightarrow 3y + y^2 - 28 = 0$
$\Rightarrow y^2 + 3y - 28 = 0$
$\Rightarrow y^2 + 7y - 4y - 28 = 0$
$\Rightarrow y(y + 7) - 4(y + 7) = 0$
$\Rightarrow (y - 4)(y + 7) = 0$
$\Rightarrow y = 4 or y = -7$ (rejected being natural no.)
When $y = 4, x = 3 + 4 = 7$ [From (ii)]
Number are $7, 4$
View full question & answer→Question 143 Marks
Find the consecutive even integers whose squares have the sum $340.$
AnswerLet the consecutive even integers be $2x$ and $2x + 2$
Then according to the given hypothesis,
$(2x)^2 + (2x + 2)^2 = 340$
$4x^2 + 4x^2 + 8x + 4 - 340 = 0$
$\Rightarrow 8x^2 + 8x - 336 = 0$
$\Rightarrow x^2 + x - 42 = 0$
$\Rightarrow x^2 + 7x - 6x - 42 = 0$
$\Rightarrow x(x + 7) - 6(x + 7) = 0$
$\Rightarrow (x + 7)(x - 6) = 0$
$\Rightarrow x = -7$ or $x = 6$
Considering, the positive integers of $x = 6$
$\Rightarrow 2x = 12$ and $2x + 2 = 14$
$\therefore$ The two consecutive even integers are $12$ and $14$
View full question & answer→Question 153 Marks
Determine the set of values of k for which the given following quadratic has real root:
$x^2 - kx + 9 = 0$
Answer$x^2 - kx + 9 = 0$
Here, $a = 1, b = -k, c = 9$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (-k)^2 - 4 \times 1 \times 9$
$= k^2 - 36$
$\because$ Roots are real
$\therefore\text{D}\geq0$
$\Rightarrow\text{k}^2-36\geq0$
$\Rightarrow\text{k}^2\geq36$
$\Rightarrow\text{k}^2\geq(\pm6)^2$
$\therefore\text{k}\geq6$ or $\text{k}\leq-6$
View full question & answer→Question 163 Marks
Determine the nature of the root of following quadratic equation:
$(x - 2a)(x - 2b) = 4ab$
Answer$(x - 2a)(x - 2b) = 4ab$
$\Rightarrow x^2 - 2bx - 2ax + 4ab - 4ab = 0$
$\Rightarrow x^2 - 2(a + b)x = 0$
Here $a = 1, b = -2(a + b), c = 0$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= {-2(a + b)}^2 - 4 \times 1 \times 0$
$= {-2(a + b)}^2$^
$\because$ D > 0
$\therefore$ Roots are real and distinct.
View full question & answer→Question 173 Marks
The sum of two numbers is $9$. The sum of their reciprocals is $\frac{1}{2}.$ Find the numbers.
AnswerGiven that the sum of two number is 9 let the two numbers be x and 9 - x
By the given hypothesis, 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 or x = 3$
$\therefore$ The two numbers are 3 and 6
View full question & answer→Question 183 Marks
Determine the set of values of k for which the given following quadratic has real root:
$2 x^2+k x-4=0$
Answer$2 x^2+k x-4=0$
Here, $a=2, b=k, c=-4$
$\therefore$ $\therefore \text { Discriminant }(D)=b^2-4 a c$
$=k^2-4 \times 2 \times(-4)$
$=k^2+32$
$\because$ The roots are real
$\therefore\text{D}\geq0$
$\Rightarrow\text{k}^2+32\geq0$
$\because\text{k}^2+32\geq0$ for all value of $\text{k }\in\text{ R}$
View full question & answer→Question 193 Marks
The sum of the squares of two numbers is $233$ and one of the number is $3$ less than twice the other number. Find the numbers.
AnswerLet the number be x
Then the other number = 2x - 3
According to the given hypothesis,
$\Rightarrow x^2 + (2x - 3)^2 = 233$
$\Rightarrow x^2 + 4x^2 + 9 - 12x = 233$
$\Rightarrow 5x^2 - 12x - 224 = 0 .....(i)$
The value of 'x' can obtained by the formula $\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
Here, $a = 5, b = -12$ and $c = -224$ from $(i)$
$\Rightarrow\text{x}=\frac{-(-12)+\sqrt{144+20\times224}}{10}=8$
$\Rightarrow\text{x}=\frac{-(-12)-\sqrt{144+20\times224}}{10}=\frac{-28}{5}$
Considering the value of x = 8
$\Rightarrow 2x - 3$
$\Rightarrow 2 \times 8 - 3$
$\Rightarrow 16 - 3$
$\Rightarrow 13$
$\therefore$ The two number are $8$ and $13.$
View full question & answer→Question 203 Marks
Find the values of k for which the given quadratic equation has real and distinct root:
$kx^2 + 6x + 1 = 0$
Answer$kx^2 + 6x + 1 = 0$
Here, $a = k, b = 6, c = 1$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (6)^2 - 4 \times k \times 1$
$= 36 - 4k$
$\because$ Roots are real and distinct,
$\therefore$ $D > 0$
$\Rightarrow 36 - 4k > 0$
$\Rightarrow 9 - k > 0$
$\Rightarrow 9 > k$
$\Rightarrow k < 9$
$\therefore k < 9$
View full question & answer→Question 213 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$3x^2 - 5x + 2 = 0$
Answer$3x^2 - 5x + 2 = 0$
Here, $a = 3, b = -5, c = 2$
$D = b^2 - 4ac$
$= (-5)^2 - 4 \times 3 \times 2$
$= 25 - 24$
$= 1$
$\therefore$ D > 0
$\therefore$ Roots are real
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-5)\pm\sqrt{1}}{2\times3}=\frac{5\pm1}{6}$
$\therefore\text{x}=\frac{5+1}{6}=\frac{6}{6}=1$
or $\text{x}=\frac{5-1}{6}=\frac{4}{6}=\frac{2}{3}$
$\text{x}=1,\frac{2}{3}$
View full question & answer→Question 223 Marks
Find two natural numbers which differ by $3$ and whose squares have the sum $117$.
AnswerLet first number $= x$
Then second number $= x - 3$
According to the condition,
$x^2 + (x - 3)^2 = 117$
$\Rightarrow x^2 + x^2 - 6x + 9 = 117$
$\Rightarrow 2x^2 - 6x + 9 - 117 = 0$
$\Rightarrow 2x^2 - 6x - 108 = 0$
$\Rightarrow x^2 - 3x - 54 = 0 (Dividing by 2)$
$\Rightarrow x^2 - 9x + 6x - 54 = 0$
$\Rightarrow x(x - 9) + 6(x - 9) = 0$
$\Rightarrow (x - 9)(x + 6) = 0$
Either x - 9 = 0, then x = 9
Or $x + 6 = 0$, then $x = -6$ which is not a natural number,
First natural number $= 9$
and second number $= 9 - 3 = 6$
View full question & answer→Question 233 Marks
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $20$ years ago, the product of their ages in years was $48$.
AnswerSum of ages of two friends = 20 years
Let present age of one friend = x years
Age of second friend = (20 - x) years
4 years ago,
Age of first friend $= x - 4$
and age of second friend $= 20 - x - 4 = 16 - x$
According to the condition,
$(x - 4)(16 - x) = 48$
$\Rightarrow 16x - x^2 - 64 + 4x = 48$
$\Rightarrow -x^2 + 20x - 64 - 48 = 0$
$\Rightarrow -x^2 + 20x - 112 = 0$
$\Rightarrow x^2 - 20x + 112 = 0$
Here a = 1, b = -20, c = 112
Discriminan $(D) = b^2 - 4ac$
$= (-20)^2 - 4 \times 1 \times 112$
$= 400 - 448 = -48$
$\therefore$ D < 0
Roots are not real
It is not possible.
View full question & answer→Question 243 Marks
If 1 is a root of the quadratic equation $3 x^2+a x-2=0$ and the quadratic equation $a\left(x^2+6 x\right)-b=0$ has equal roots, find the value of $b \mid$
Answer$1$ is one root of $3x^2 + ax - 2 = 0$
$\therefore$ $3(1)^2 + a \times 1 - 2 = 0$
$\Rightarrow 3 + a - 2 = 0$
$\Rightarrow a + 1 = 0$
$\Rightarrow a = -1$
Now in equation $a(x^2 + 6x) - b = 0$
$\Rightarrow -1(x^2 + 6x) - b = 0$
$\Rightarrow -x^2 - 6x - b = 0$
$\Rightarrow x^2 + 6x + b = 0$
Here $A = 1, B = 6, C = b$
$\therefore$ $D = B^2 - 4AC$
$\Rightarrow D = (6)^2 - 4 \times 1 \times k$
$\Rightarrow D = 36 - 4k$
$\because$ Roots are equal
$\Rightarrow D or B^2 - 4AC = 0$
$\Rightarrow 36 - 4k = 0$
$\Rightarrow 4k = -36$
$\Rightarrow\text{k}=\frac{-36}{4}=-9$
View full question & answer→Question 253 Marks
Solve the following quadratic equations by factorization:
$\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}$ $(\text{x}\neq1,-5)$
$\frac{\text{x}+5-\text{x}+1}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{6}{(\text{x}-1)(\text{x}+5)}=\frac{6}{7}$
$\Rightarrow\frac{1}{(\text{x}-1)(\text{x}+5)}=\frac{1}{7}$ (Dividing by 6)
$(x - 1)(x + 5) = 7$
$\Rightarrow x^2 + 5x - x - 5 = 7$
$\begin{Bmatrix}\because-12=6\times(-2)\\+4=6-2\end{Bmatrix}$
$\Rightarrow x^2 + 4x - 5 - 7 = 0$
$\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$
Either $x + 6 = 0$, then $x = -6$
or $x - 2 = 0$, then $x = 2$
$\therefore$ $x = 2, -6$
View full question & answer→Question 263 Marks
Find the value of k for which the following equations have real and equal roots:
$x^2 + k(2x + k - 1) + 2 = 0$
AnswerThe given equation is $x^2 + k(2x + k - 1) + 2 = 0$
$\Rightarrow x^2 + 2kx + k(k - 1) + 2 = 0$
So, $a = 1, b = 2k, c = k(k - 1) + 2$
We know $D = b^2 - 4ac$
$\Rightarrow D = (2k)^2 - 4 \times 1 \times [k(k - 1) + 2]$
$\Rightarrow D = 4k^2 - 4[k^2 - k + 2]$
$\Rightarrow D = 4k^2 - 4k2 + 4k - 8$
$\Rightarrow D = 4k - 8$
$\Rightarrow D = 4(k - 2)$
For equal roots, $D = 0$
Thus, $4(k - 2) = 0$
So, $k = 2$
View full question & answer→Question 273 Marks
Find the value of k for which the root are real and equal in the following equations:
$\text{kx}^2-2\sqrt{5}\text{x}+4=0$
AnswerThe given equation $\text{kx}^2-2\sqrt{5}\text{x}+4=0$ is in the from of $ax^2 + bx + c = 0$ where
$\text{a}=\text{k},\text{b}=-2\sqrt{3}$ and $\text{c}=4$
Gievn that, the equation has and equal roots,
i.e., $\text{D}=\text{b}^2-4\text{ac}=0$
$\Rightarrow(-2\sqrt{5})^2-4\times\text{k}\times4=0$
$\Rightarrow20=16\text{k}$
$\Rightarrow\text{k}=\frac{20}{16}=\frac{5}{4}$
$\therefore\text{k}=\frac{5}{4}$
$\therefore$ The value of $\text{k}=\frac{5}{4}$
View full question & answer→Question 283 Marks
Find the value of k for which root are real and equal in the following equations:
$(k + 1)x^2 + 2(k + 3)x + (k + 8) = 0$
AnswerThe given quadric equation is $(k + 1)x^2 + 2(k + 3)x + (k + 8) = 0$, and roots are real and equal
Then find the value of k.
Here, a = (k + 1), b = 2(k + 3) and c = k + 8
As we know that $d = b^2 - 4ac$
Putting the value of a $= (k + 1), b = 2(k + 3) and c = k + 8$
$= (2(k + 3))^2 - 4 \times (k + 1) \times (k + 8)$
$= (4k^2 + 24k + 36) - 4(k^2 + 9k + 8)$
$= 4k^2 + 24k + 36 - 4k^2 - 36k - 32$
$= -12k + 4$
The given equation will have real and equal roots, if $D = 0$
$-12k + 4 = 0$
$\text{k}=\frac{4}{12}$
$\text{k}=\frac{1}{3}$
Therefore, the value of $\text{k}=\frac{1}{3}$
View full question & answer→Question 293 Marks
In the following, find the value of k for which the given value is a solution of the given equation:
$\text{kx}^2+\sqrt{2}\text{x}-4=0,\text{x}=\sqrt{2}$
Answer$\text{kx}^2+\sqrt{2}\text{x}-4=0,\text{x}=\sqrt{2}$
Given that $\text{x}=\sqrt{2}$ is a root at the given equation
$\text{kx}^2+\sqrt{2}\text{x}-4=0$
$\Rightarrow\text{x}=\sqrt{2}$ Satisfies the equation
i.e., $\text{k}(\sqrt{2})^2+\sqrt{2}(\sqrt{2})-4=0$
$\Rightarrow2\text{k}+2-4=0$
$\Rightarrow2\text{k}-2=0$
$\Rightarrow2\text{k}=2$
$\Rightarrow\text{k}=1$
View full question & answer→Question 303 Marks
The height of a right triangle is $7\ cm$ less than its base. If the hypotenuse is $13\ cm$, form the quadratic equation to find the base of the triangle.
AnswerGiven that in a right triangle is 7cm less than its base
Let base of the triangle be denoted by x
⇒ Height of the triangle = (x - 7)cm
We have hypotenuse of the triangle = 13cm
We know taht, in a right triangle
$(base)^2+ (Height)^2 = (Hypotenuse)^2$
$\Rightarrow (x)^2 + (x - 7)^2 = (13)^2$
$\Rightarrow x^2 + x^2 - 14x + 49 = 169$
$\Rightarrow 2x^2 - 14x + 49 - 169 = 0$
$\Rightarrow 2x^2 - 14x - 120 = 0$
$\Rightarrow 2(x^2- 7x - 60) = 0$
$\Rightarrow x^2- 7x - 60 = 0$
$\therefore$ The required quadratic equation is $x^2 - 7x - 60 = 0$
View full question & answer→Question 313 Marks
Ashu is x years old while his mother Mrs. Veena is $x^2$ years old. Five years hence Mrs. Veena will be three times old as Ashu. Find their present ages.
AnswerGiven that, Ashu is x years old while his mother Mrs. Veena is $x^2$ years old.
Ashu’s present age = x years and Mrs. Veena’s present age $= x^2$ years
And also given that, after 5 years Mrs. Veena will be three times old as Ashu.
Ashu's age after 5 years $= (x + 5)$ years
And mrs. veena's age after $5$ years $= (x^2 + 5)$ years
But given that,
$\Rightarrow (x^2 + 5) = 3(x + 5)$
$\Rightarrow x^2+ 5 = 3x + 15$
$\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 or x = -2$
Since, age cannot be in negative values. So x = 5 years.
$\therefore$ Present age of Ashu is $x = 5$ years and Present age of Mrs. Veena is $x^2 = 5^2$ years
$\Rightarrow 25$ years.
View full question & answer→Question 323 Marks
Two number differ by $3$ and their product is $504$. Find the numbers.
AnswerLet the two numbers be x and $x - 3$ given that $x(x - 3) = 504$
$\Rightarrow x^2 - 3x - 504 = 0$
$\Rightarrow x^2 - 24x + 21x - 504 = 0$
$\Rightarrow x(x - 24) + 21(x - 24) = 0$
$\Rightarrow (x - 24) + 21(x - 24) = 0$
$\Rightarrow (x - 24)(x + 21) = 0$
$\Rightarrow x = 24 or x = 21$
Case I: If $x - 3 = 24 - 3 = 21$
Case II: If $x = 21, x = 3 = 24$
$\therefore$ The two numbers are $21, 24$ or $-21, -24$
View full question & answer→Question 333 Marks
The sum of two numbers is $8$ and $15$ times the sum of their reciprocals is also $8$. Find the numbers.
AnswerSum of two nubers $= 8$
Let first number $= x$
Then second number $= 8 - x$
According to the condition,
$15\Big[\frac{1}{\text{x}}+\frac{1}{8-\text{x}}\Big]=8$
$\Rightarrow\frac{1}{\text{x}}+\frac{1}{8-\text{x}}=\frac{8}{15}$
$\Rightarrow\frac{8-\text{x}+\text{x}}{\text{x}(8-\text{x})}=\frac{8}{15}$
$\Rightarrow\frac{8}{\text{x}(8-\text{x})}=\frac{8}{15}$
$\Rightarrow x(8 - x) = 15$
$\Rightarrow 8x - x^2 - 15 = 0$
$\Rightarrow -x^2 + 8x - 15 = 0$
$\Rightarrow x^2 - 8x + 15 = 0$
$\Rightarrow x^2 - 3x - 5x + 15 = 0$
$\begin{Bmatrix}\because15=-3\times(-5)\\-8=-3-5 \end{Bmatrix}$
$\Rightarrow x(x - 3) - 5(x - 3) = 0$
$\Rightarrow (x - 3)(x - 5) = 0$
Either $x - 3 = 0$, then $x = 3$
Or $x - 5 = 0$, then $x = 5$
- If $x = 3$, then First number$ = 3$ and second number $= 8 - 3 = 5$
- If $x = 5$, then First number $= 5$ and second number $= 8 - 5 = 3$
Numbers are 3, 5 View full question & answer→Question 343 Marks
The product of Shikha's age five years ago and her age $8$ years later is $30$, her age at both times being given in years. Find her present age.
AnswerLet present age of Shikha $= x$ years
5 years ago, her age was $= (x - 5)$ years
and 8 years later, her age will be $= (x + 8)$ years
According to the condition,
$(x - 5)(x + 8) = 30$
$\Rightarrow x^2 + 3x - 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$
Either$ x + 1 0 = 0$, then $x = -10$ which is not possible being negative
Or $x - 7 = 0,$ then $x = 7$
Her present age $= 7$ years.
View full question & answer→Question 353 Marks
Solve the following quadratic equation by factorization:
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
AnswerWe have
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$[-1 \times a^2b^2 = -a^2b^2 \Rightarrow -a^2b^2 = -a^2 \times b^2 = -a^2 \times b^2]$
$\Rightarrow 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{1}{\text{a}^2}$ and $\text{x}= \frac{1}{\text{b}^2}$ are the two root of the given equation.
View full question & answer→Question 363 Marks
The sum of the squares of two consecutive even numbers is 340 . Find the numbers.
AnswerLet first even number $=2 x$
Then second number $=2 x+2$
$\Rightarrow (2x)^2 + (2x + 2)^2 = 340$
$\Rightarrow 4x^2 + 4x^2 + 8x + 4 - 340 = 0$
$\Rightarrow 8x^2 + 8x - 336 = 0$
$\Rightarrow x^2+ x - 42 = 0 (Dividing by 8)$
$\Rightarrow x^2 + 7x - 6x - 42 = 0$
$\Rightarrow x(x + 7) - 6(x + 7) = 0$
$\Rightarrow (x + 7)(x - 6) = 0$
Either $x+7=0$, then $x=-7$ but not possible being negative
or $x-6=0$, then $x=6$
Number are 12,14
View full question & answer→Question 373 Marks
Solve the following quadratic equations by factorization:
$2x^2 + ax - a^2 = 0$
Answer$2x^2 + ax - a^2 = 02x^2 + 2ax - ax - a^2 = 0$
$2x(x + a) - a(x + a) = 0$
$(x + a)(2x - a) = 0$
$x + a = 0 or 2x - a = 0$
$x = -a$ or $\text{x}=\frac{\text{a}}{2}$
Alternate Answer
First calculate $D=b^2-4 a c$
Then apply $x =\frac{- b \pm \sqrt{ D }}{2 a }$
We get $x=-a, x=\frac{a}{2}$
View full question & answer→Question 383 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$3\text{a}^2\text{x}^2+8\text{abx}+4\text{b}^2=0,$ $\text{a}\neq0$
Answer$3\text{a}^2\text{x}^2+8\text{abx}+4\text{b}^2=0,$
$\text{a}\neq0$The given equation in the form of $ax^2 + bx + c = 0$
Here, $a = 3a^2, b = 8ab, c = 4b^2$ $[$given $\text{a}\neq0]$
$D = b^2 - 4ac$
$= (8ab)^2 - 4 \times 3a^2 \times 4b^2$
$= 64a^2b^2 - 48a^2b^2$
$= 16a^2b^2 > 0$
As Q = 0, the given equation has real roots, 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{-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{-2\text{b}}{\text{a}}$
$\therefore$ The roots of the given equation are $\frac{-2\text{b}}{3\text{a}},\frac{-2\text{b}}{\text{a}}$
View full question & answer→Question 393 Marks
Find the value of k for which root are real and equal in the following equations:
$9x^2 - 24x + k = 0$
AnswerThe given equation is $9x^2 - 24x + k = 0$
This equation is in form of $ax^2 + bx + c = 0$
Here, a = 9, b = -24 and c = k
Given that, the nature of the roots of this equation is real and equal
i.e., $D = b^2 - 4ac = 0$
$\Rightarrow (-24)^2 - 4 \times 9 \times k = 0$
$\Rightarrow 576 - 36k = 0$
$\Rightarrow\text{k}=\frac{576}{36}$
$\therefore$ k = 16
$\therefore$ The value of k = 16
View full question & answer→Question 403 Marks
In the following, find the value of k for which the given value is a solution of the given equation:
$7\text{x}^2+\text{kx}-3=0,$ $\text{x}=\frac{2}{3}$
AnswerWe are given here that,
$7\text{x}^2+\text{kx}-3=0,$ $\text{x}=\frac{2}{3}$
Now, as we know that $\text{x}=\frac{2}{3}$ is a solution of the quadratic equation, hence it should satify the equation. Therefore substituting $\text{x}=\frac{2}{3}$ in the above equation gives us,
$7\Big(\frac{2}{3}\Big)^2+\text{k}\Big(\frac{2}{3}\Big)-3=0$
$\frac{28+6\text{k}-27}{9}=0$
$6\text{k}=-1$
$\text{k}=-\frac{1}{6}$
Hence, the value of $\text{k}=-\frac{1}{6}$
View full question & answer→Question 413 Marks
A fast train takes one hour less than a slow train for a journey of 200 km . If the speed of the slow train is $10 km / hr$ less than that of the fast train, find the speed of the two trains.
AnswerTotal journey $=200 km$
Let the speed of fast train $=x km / hr$
Then speed of slow train $=(x-10) km / hr$
According to the condition,
$\frac{200}{\text{x}-10}-\frac{200}{\text{x}}=1$
$\Rightarrow\frac{200\text{x}-200\text{x}+2000}{\text{x}(\text{x}-10)}=1$
$\Rightarrow\frac{2000}{\text{x}^2-10\text{x}}=1$
$\Rightarrow x^2 - 10x = 2000$
$\Rightarrow x^2 - 10x - 2000 = 0$
$\Rightarrow x^2 - 50x + 40x - 2000 = 0$
$\begin{cases}\because-2000=-50\times40\\-10=-50+40\end{cases}$
$\Rightarrow x(x - 50) + 40 (x - 50) = 0$
$\Rightarrow (x - 50) (x + 40) = 0$
Either $x-50=0$, then $x=50$
or $x+40=0$, then $x=-40$ but it is not possible being negative
Speed of the fast trin $=50 km / hr$
and speed of the slow trin $=50-10=40 km / hr$
View full question & answer→Question 423 Marks
In the following, determine the set of values of k for which the given quadratic equation has real root:
$3x^2 + 2x + k = 0$
Answer$3x^2 + 2x + k = 0$
Here $a = 3, b = 2, c = k$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (2)^2 - 4 \times 3 \times k$
$= 4 - 12k$
$\because$ Roots are real
$\therefore\text{D}\geq0$
$\Rightarrow4-12\text{k}\geq0$
$\Rightarrow4\geq12\text{k}$
$\Rightarrow12\text{k}\leq4$
$\Rightarrow\text{k}\leq\frac{4}{12}$
$\Rightarrow\text{k}\leq\frac{1}{3}$
View full question & answer→Question 433 Marks
In the following, determine the set of values of k for which the given quadratic equation has real root:
$kx^2 + 6x + 1 = 0$
AnswerThe given equation is $k x^2+6 x+1=0$, and roots are real.
Then find the value of $k$.
Here, $a=k, b=6$ and $c=1$
As we know that $D=b^2-4 a c$
Putting the value of $a=k, b=6$ and $c=1$
$= (6)^2 - 4 \times k \times 1$
$= 36 - 4k$
The given equation will have real roots, if $\text{D}\geq0$
$36-4\text{k}\geq0$
$4\text{k}\leq36$
$\text{k}\leq\frac{36}{4}$
$\text{k}\leq9$
Therefore, the value of $\text{k}\leq9$
View full question & answer→Question 443 Marks
Solve the following quadratic equations by factorization:
$48x^2 - 13x - 1 = 0$
Answer$48x^2 - 13x - 1 = 0$
$\Rightarrow 48x^2 - 16x + 3x - 1 = 0$
$\begin{cases}\because48\times(-1)=48\\\therefore-48=-16\times3\\-13=-16+3\end{cases}$
$\Rightarrow 16x(3x - 1) + 1(3x - 1) = 0$
$\Rightarrow (3x - 1)(16x + 1) = 0$
Either $3 x-1=0$,
Then $3 x=1$
$\Rightarrow\text{x}=\frac{1}{3}$
Or $16 x+1=0$,
Then $16 x=-1$
$\Rightarrow\text{x}=\frac{-1}{16}$
Roots are $\text{x}=\frac{1}{3},\frac{-1}{16}$
View full question & answer→Question 453 Marks
The sum of a number and its positive square root is $\frac{6}{25}.$ Find the number.
AnswerLet the number be x
By the hypothesis, we have
$\Rightarrow\text{x}+\sqrt{\text{x}}=\frac{6}{25}$
Let us assume that $x = y^2$, we get
$\Rightarrow\text{y}^2+\text{y}=\frac{6}{25}$
$\Rightarrow25\text{y}^2+25\text{y}-6 = 0$
The value of 'y' can be obtaines by $\text{y} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
Where $a = 25, b = 25, c = -6$
$\Rightarrow\text{y}=\frac{-25\pm\sqrt{625+600}}{50}$
$\Rightarrow\text{y}=\frac{-25\pm35}{50}$
$\Rightarrow\text{y}=\frac{1}{5}$ or $\frac{-11}{10}$
$\Rightarrow\text{x}=\text{y}^2$
$\Rightarrow\text{x}=\Big(\frac{1}{5}\Big)^2$
$\Rightarrow\text{x}=\frac{1}{25}$
$\therefore$ The number $\text{x}=\frac{1}{25}$
View full question & answer→Question 463 Marks
In the following, determine whether the given quadratic equation have real roots and if so, find the roots:
$\sqrt{3}\text{x}^2+10\text{x}-8\sqrt{3}=0$
Answer$\sqrt{3}\text{x}^2+10\text{x}-8\sqrt{3}=0$
The given equation is in the form of $ax^2 + bx + c = 0$
Here $\text{a}=\sqrt{3},\text{b}=10$ and $\text{c}=-8\sqrt{3}$
$\text{D}=\text{b}^2-4\text{ac}$
$=(10)^2-4\times\sqrt{3}\times-8\sqrt{3}$
$=100+96=196$
AS Q > 0, the given equation has real roota, given by
$\alpha=\frac{-\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-10+\sqrt{196}}{2\times\sqrt{3}}$
$=\frac{2\sqrt{3}}{3}=\frac{2}{\sqrt{3}}$
$\beta=\frac{-\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-10-\sqrt{196}}{2\times\sqrt{3}}$
$=-4\sqrt{3}$
$\therefore$ The roots of the equation are $\frac{2}{\sqrt{3}}$ and $-4\sqrt{3}$
View full question & answer→Question 473 Marks
Find the values of k for which the given quadratic equation has real and distinct root:
$x^2 - kx + 9 = 0$
AnswerThe given quadric equation is $x ^2- kx +9=0$, and roots are real and distinct.
Then find the value of $k$.
$A=1, b=-k \text { and } c=9$
As we know that $D=b^2-4 a c$
Putting the value of $a=1, b=-k$ and $c=9$
$\Rightarrow D=(k)^2-4 \times 1 \times 9$
$\Rightarrow D=K^2-36$
The given equation will have real and distinct roots, if $D>0$
$\Rightarrow k^2-36>0$
Now factorizing of the above equation
$\Rightarrow k^2-36>0$
$\Rightarrow k^2>36$
$\Rightarrow k>\sqrt{36}$
$\Rightarrow k= \pm 6$
$\Rightarrow k<-6 \text { or } k>6$
Therefore, the value of $k<-6$ or, $k>6$
View full question & answer→Question 483 Marks
Is there any value of ' $a$ ' for which the equation $x^2+2 x+\left(a^2+1\right)=0$ has real root?
AnswerLet quadratic equation $x^2+2 x+\left(a^2+1\right)=0$
Here $a =1, b=2$ and $c =\left( a ^2+1\right)$
As we know that $D=b^2-4 a c$
Putting the value of $a=1, b=2$ and $c=\left(a^2+1\right)$
$\Rightarrow D = (2)^2 - 4 \times 1 \times (a^2 + 1)$
$\Rightarrow D = 4 - 4(a^2 + 1)$
$\Rightarrow D = -4a^2$
The given equation will have equal roots, id $D>0$
i.e.,$ -4a^2 > 0$
$\Rightarrow a^2 < 0$
Which is not possible, as the square of any number is always positive.
Thus, No, there is no any real value of a which the given equation has real roots.
View full question & answer→Question 493 Marks
Solve the following quadratic equations by factorization:
$\text{x}^2+\Big(\text{a}+\frac{1}{\text{a}}\Big)\text{x}+1=0$
AnswerWe have been given
$\text{x}^2+\Big(\text{a}+\frac{1}{\text{a}}\Big)\text{x}+1=0$
Therefore,
$\text{x}^2+\text{ax}+\frac{1}{\text{a}}\text{x}+1=0$
$\text{x}(\text{x}+\text{a})+\frac{1}{\text{a}}(\text{x}+\text{a})=0$
$\Big(\text{x}+\frac{1}{\text{a}}\Big)(\text{x}+\text{a})=0$
Therefore,
$\text{x}+\frac{1}{\text{a}}=0$
$\text{x}=-\frac{1}{\text{a}}$
or, $\text{x}+\text{a}=0$
$\text{x}=-\text{a}$
Hence, $\text{x}=-\frac{1}{\text{a}}$ or $\text{x}=-\text{a}$
View full question & answer→Question 503 Marks
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3,$ $\text{x}\neq0,2$
Answer$\frac{1}{\text{x}}-\frac{1}{\text{x}-2}=3$ $(\text{x}\neq0,2)$
$\frac{\text{x}-2-\text{x}}{\text{x}(\text{x}-2)}=3$
$\Rightarrow\frac{-2}{\text{x}^2-2\text{x}}=3$
$\Rightarrow 3x^2 - 6x = -2$
$\Rightarrow 3x^2 - 6x + 2 = 0$
Here $a = 3, b = -6, c = 2$
$D = b^2 - 4ac = (-6)^2 - 4 \times 3 \times 2$
$= 36 - 24 = 12$
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\text{x}={-(-6) \pm \sqrt{12} \over 2\times3}$
$\text{x}={6 \pm2 \sqrt{3} \over6}$
$\text{x}=\frac{2(3\pm\sqrt{3})}{6}$
$\text{x}=\frac{3\pm\sqrt{3}}{3}$
View full question & answer→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$ Discriminant $(D) = b^2 - 4ac$
$= [-2(k - 1)]^2 - 4(k + 1) \times 1$
$= 4(k^2 - 2k + 1) - 4(k + 1)$
$= 4k^2 - 8k + 4 - 4k - 4$
$= 4k^2 - 12k$
$\because$ Roots are real and equal
$\therefore$ $D = 0$
$4k^2- 12k = 0$
$k^2 - 3k = 0$ (Dividing by $4$)
$k(k - 3) = 0$
Either$ k = 0$
$or k - 3 = 0$, then$ k = 3$
$\therefore$ $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:
$kx^2 + 4x + 1 = 0$
Answer$kx^2 + 4x + 1 = 0$
Here $a = k, b = 4, c = 1$
$\therefore$ Discriminant $(D) = b^2 - 4ac = (4)^2 - 4 \times k \times 1 = 16 - 4k$
$\because$ Roots are real and equal
$\therefore$ $D = 0$
$\Rightarrow 16 - 4k = 0$
$\Rightarrow 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,
$4x^2 + 4 = 17x$
$4x^2 - 17x + 4 = 0$
$4x^2 - 17x + 4 = 0$
$4x^2 - x - 16x + 4 = 0$
$x(4x - 1) - 4(4x - 1) = 0$
$(4x - 1)(x - 4) = 0$
$(4x - 1) = 0$
$\text{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$
$\Rightarrow x^2 + {(a) - (a + 1)}x - a(a + 1) = 0$
${ 1 = a - a + 1}$
$\Rightarrow x^2 + ax - (a + 1)x - a(a + 1) = 0$
$\Rightarrow x(x + a) - (a + 1)(x + a) = 0$
$\Rightarrow (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:
$5x^2 - 3x - 2 = 0$
AnswerWe have,$5x^2 - 3x - 2 = 0$
$\Rightarrow 5x^2 - 5x + 2x - 2 = 0$
$\Rightarrow 5x(x - 1) + 2(x - 1) = 0$
$\Rightarrow (x - 1)(5x + 2) = 0$
$\Rightarrow (x - 1) = 0$ or $5x + 2 = 0$
$\Rightarrow 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 + 9x - 5x - 45 = 15$
$\Rightarrow x^2 + 4x - 60 = 0$
$\Rightarrow x^2 + 10x - 6x - 60 = 0$
$\Rightarrow x(x + 10) - 6(x + 10) = 0$
$\Rightarrow (x + 10)(x - 6) = 0$
$\Rightarrow x + 10 = 0 or x - 6 = 0$
$\Rightarrow x = -10 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$
$A.T.Q., (x)^2 + (x + 2)^2 = 970$
$\Rightarrow x^2 + x^2 + 4x + 4 - 970 = 0$
$\Rightarrow 2x^2 + 4x - 966 = 0$
$\Rightarrow x^2 + 2x - 483 = 0$
$\Rightarrow x^2 + 23x - 21x - 483 = 0$
$\Rightarrow x(x + 23) - 21(x + 23) = 0$
$\Rightarrow (x - 21)(x + 23) = 0$
Either $x - 21 = 0 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 (7x)^2 + [7(x + 1)]^2 = 637$
$\Rightarrow 49x^2 + 49(x^2 + 2x + 1) = 637$
$\Rightarrow 49x^2 + 49x^2 + 98x + 49 - 637 = 0$
$\Rightarrow 98x^2 + 98x - 588 = 0$
$\Rightarrow x^2 + x - 6 = 0$
$\Rightarrow x^2 + 3x - 2x - 6 = 0$
$\Rightarrow x(x + 3) - 2(x + 3) = 0$
$\Rightarrow (x - 2)(x + 3) = 0$
$\Rightarrow x - 2 = 0 or x + 3 = 0$
$\Rightarrow x = 2 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
$5x \times 5(5x + 1) = 300$
$\Rightarrow 25x(x + 1) = 300$
$\Rightarrow x^2 + x = 12$
$\Rightarrow x^2 + x - 12 = 0$
$\Rightarrow x^2 + 4x - 3x - 12 = 0$
$\Rightarrow x(x + 4) - 3(x + 4) = 0$
$\Rightarrow (x + 4)(x - 3) = 0$
$\Rightarrow x = -4 or x = 3$
$If x = -4, 5x = -20, 5(x + 1) = -15$
$x = 3, 5x = 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 - 3x + 2 = 0, x = 2, x = -1$
Answer$x^2 - 3x + 2 = 0. x = 2, x = -1$
$$Here, $L.H.S. = x^2 - 3x + 2$ 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 $\neq$ 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:
$2kx^2 - 40x + 25 = 0$
AnswerThe given quadric equation is $2kx^2 - 40x + 25 = 0$, 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 $c = 25$
$= (-40)^2 - 4 \times 2k \times 25$
$= 1600 - 200k$
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,
$(x - 5)(45 - x - 5) = 128$
$(x - 5)(40 - x) = 128$
$40x - x^2 - 200 + 5x - 128 = 0$
$x^2 - 45x + 200 + 128 = 0$
$x^2 - 45x + 328 = 0$
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,
$(x)^2 + (x + 1)(x + 2) = 46$
$\Rightarrow x^2 + x^2 + 3x + 2 = 46$
$\Rightarrow 2x^2 + 3x + 2 - 46 = 0$
$\Rightarrow 2x^2 + 3x - 44 = 0$
$\Rightarrow 2x^2 + 11x - 8x - 44 = 0$
$\Rightarrow x(2x + 11) - 4(2x + 11) = 0$
$\Rightarrow (2x + 11)(x - 4) = 0$
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,
$\frac{1}{\text{x}}+\frac{1}{18-\text{x}}=\frac{1}{4}$
$\Rightarrow\frac{18-\text{x}+\text{x}}{\text{x}(18-\text{x})}=\frac{1}{4}$
$\Rightarrow\frac{18}{18\text{x}-\text{x}^2}=\frac{1}{4}$
$72 = 18x - x^2$
$x^2 - 18x + 72 = 0$
$\Rightarrow x^2 - 12x - 6x + 72 = 0$
$\Rightarrow x(x - 12) - 6(x - 12) = 0$
$\Rightarrow (x - 12)(x - 6) = 0$
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:
$abx^2 + (b^2 - ac)x - bc = 0$
AnswerWe have,
$abx^2 + (b^2 - ac)x - bc = 0$
$[abx - bc = -ab^2c$
$\Rightarrow -ab^2c = b^2 \times -ac and b^2- ac = b^2 + (-ac)]$
$\Rightarrow abx^2 + b^2x - acx - bc = 0$
$\Rightarrow bx(ax + b) - c(ax + b) = 0$
$\Rightarrow (ax + b)(bx - c) = 0$
$\Rightarrow ax + b = 0 or bx - c = 0$
$\Rightarrow\text{x}=-\frac{\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{c}}{\text{b}}$
$\therefore\text{x}=-\frac{\text{b}}{\text{a}}$ 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 $x^2 - kx + 4 = 0$ has equal roots.
AnswerThe given quadric equation is $x^2 - kx + 4 = 0$, and roots are equal.
Then find the value of k.
Here, $a = 1, b = -k$ and $c = 4$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 1, b = -k$ and $c = 4$
$= (-k)^2 - 4 \times 1 \times 4$
$= k2 - 16$
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
According to question $x^2 - y^2 = 180$ and $y^2 = 8x$
$\Rightarrow x^2 - 8x = 180$
$\Rightarrow x^2 - 8x - 180 = 0$
$\Rightarrow x^2 + 10x - 18x - 180 = 0$
$\Rightarrow x(x + 10) - 18(x + 10) = 0$
$\Rightarrow (x + 10)(x - 18) = 0$
$\Rightarrow x = -10 or x = 18$
Case I: $x = 18$
$\Rightarrow 8x = 8 \times 18 = 144$
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
$25x(x + 1) = -4$
$25x^2 + 25x + 4 = 0$
$25x^2 + 20x + 5x + 4 = 0$
$5x(5x + 4) + 1(5x + 4) = 0$
$(5x + 1)(5x + 4) = 0$
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:
$6x^2 + 11x + 3 = 0$
AnswerWe have been given
$6x^2 + 11x + 3 = 0$
$6x^2 + 9x + 2x + 3 = 0$
$3x(2x + 3) + 1(2x + 3) = 0$
$(2x + 3)(3x + 1) = 0$
$2x + 3 = 0$
$\text{x}=\frac{-3}{2}$
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 $8$
$\Rightarrow x(x - 2) = 8$
$\Rightarrow x^2 - 2x - 8 = 0$
$\Rightarrow x^2 - 4x + 2x - 8 = 0$
$\Rightarrow x(x - 4) + 2(x - 4) = 0$
$\Rightarrow (x - 4)(x + 2) = 0$
$\Rightarrow x = 4 or x = -2$
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:
$9a^2b^2x^2 - 24abcdx + 16c^2d^2 = 0$, $\text{a}\neq0,\text{b}\neq0$
AnswerThe given quadric equation is $9a^2b^2x^2 - 24abcdx + 16c^2d^2 = 0$
Here, $a = 9a^2b^2, b = -24$abcd and $c = 16c^2d^2$
As we know that $D = b^2 - 4ac$
Putting the value of $a = 9a^2b^2, b = -24abcd$ and $c = 16c^2d^2$
$= (24abcd)^2 - 4 \times 9a^2b^2 \times 16c^2d^2$
$= (576a^2b^2c^2d^2) - 576a^2b^2c^2d^2$
$= 0$
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
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3}$
$3(x^2 - 5x + 4 + x^2 - 5x + 6) = 10(x^2 - 6x + 8)$
$4x^2 - 30x + 50 = 0$
$2x^2- 15x + 25 = 0$
$2x^2- 10x - 5x + 25 = 0$
$2x(x - 5) - 5(x - 5) = 0$
$(2x - 5)(x - 5) = 0$
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,
$(2x + 1)^2 + (2x - 1)^2 = 394$
$4x^2 + 4x + 1 + 4x^2 - 4x + 1 = 394$
$8x^2 + 2 = 394$
$8x^2 = 394 - 2$
$\text{x}^2=\frac{392}{8}$
$\text{x}^2=49$
$\text{x}=\sqrt{49}$
$\text{x}=\pm7$
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:
$x^2 + 2ab = (2a + b)x$
AnswerWe have
$x^2 + 2ab = (2a + b)x$
$\Rightarrow x^2 - (2a + b)x + 2ab = 0$
[$\because$ $2ab = -8a \times -b$
$\Rightarrow -(8a + b) = -8a - b$
$\Rightarrow x^2 - 2ax - bx + 2ab = 0$
$\Rightarrow x - (x - 8a) - b(x - 2a) = 0$
$\Rightarrow (x - 8a)(x - b) = 0$
$\Rightarrow x - 8a = 0 or x - b = 0$
$\Rightarrow x = 8a = 0$ 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:
$3x^2 = -11x - 10$
Answer$3x^2 = -11x - 10$
$\Rightarrow 3x^2 + 11x + 10 = 0$
$\Rightarrow 3x^2 + 11x + 10 = 0$
$\begin{cases}\because3\times10=30\\\therefore30=5\times6\\11=5+6\end{cases}$
$\Rightarrow 3x^2 + 6x + 5x + 10 = 0$
$\Rightarrow 3x(x + 2) + 5(x + 2) = 0$
$\Rightarrow (x + 2)(3x + 5) = 0$
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:
$2x^2 + 3x + k = 0$
AnswerThe given equation is $2x^2 + 3x + k = 0$
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:
$4x^2 - 3kx + 1 = 0$
Answer$4x^2 - 3kx + 1 = 0 $Here $a = 4, b = -3k, c = 1$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= (-3k)^2 - 4 \times 4 \times 1$
$= 9k^2 - 16$
$\therefore$ Roots are real and equal
$\therefore$ $D = 0$
$\Rightarrow 9k^2 - 16 = 0$
$\Rightarrow 9k^2 = 16$
$\Rightarrow\text{k}^2=\frac{16}{9}$
$\text{k}=\Big(\pm\frac{4}{3}\Big)^2$
$\therefore\text{k}=\pm\frac{4}{3}$
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:
$16x^2 = 24x + 1$
Answer$16x^2 = 24x + 1$
$\Rightarrow 16x^2 - 24x - 1 = 0$
$\therefore$ Discriminate $= b^2 - 4ac$
$= (-24)^2 - 4 \times 16 \times (-1)$
$= 576 + 64 = 640$
$\because$ 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$\frac{1}{\text{x}}+\frac{2}{2\text{x}-3}=\frac{1}{\text{x}-2},$ $\text{x}\neq0,\frac{3}{2},2$
$\Rightarrow\frac{(2\text{x}-3)+2\text{x}}{\text{x}(2\text{x}-3)}=\frac{1}{(\text{x}-2)}$
$\Rightarrow (x - 2)(4x - 3) = x(2x - 3)$
$\Rightarrow 4x^2 - 11x + 6 = 2x^2 - 3x$
$\Rightarrow 2x^2 - 8x + 6 = 0$
$\Rightarrow x^2 - 4x + 3 = 0$
$\Rightarrow x^2 - 3x - x + 3 = 0$
$\Rightarrow x(x - 3) - (x - 3) = 0$
$\Rightarrow (x - 1)(x - 3) = 0$
$\therefore$ $x - 1 = 0$
$\Rightarrow x = 1$
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 5km/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 150km $=\frac{150}{\text{x}}\text{hr}$
Time taken by the train under increased speed to cove 150km $=\frac{150}{\text{(x+5)}}\text{hr}$
Therefore,
$\frac{150}{\text{x}}-\frac{150}{(\text{x}+5)}=1$
$\frac{\{150(\text{x+5})-150\text{x}\}}{\text{x}(\text{x}+5)}=1$
$\frac{150\text{x}+750-150\text{x}}{\text{x}^2+5\text{x}}=1$
$750 = x^2+ 5x$
$x^2 + 5x - 750 = 0$
$x^2 - 25x + 30x - 750 = 0$
$x(x - 25) + 30(x - 25) = 0$
$(x - 25)(x + 30) = 0$
So, either
$(x - 25) = 0$
$x = 25$
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 = 25km/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:
$4x^2 - 3kx + 1 = 0$
AnswerThe given quadric equation is $4x^2 - 3kx + 1 = 0$, 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 $c = 1$
$= (-3k)^2 - 4 \times 4 \times 1$
$= 9k^2 - 16$
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)
$(10x + 2x) = 2x \times 2x$
$4x^2 - 12x = 0$
$4x(x - 3) = 0$
$x(x - 3) = 0$
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
$= 10x + y$
$= 10 \times 3 + 6$
$= 36$
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
$(2x - 5)^2 - x^2 = 88$
$4x^2 - 20x + 25 - x^2 - 88 = 0$
$3x^2 - 20x - 63 = 0$
$3x^2 - 27x + 7x - 63 = 0$
$3x(x - 9) + 7(x - 9) = 0$
$(x - 9)(3x + 7) = 0$
$(x - 9) = 0$
$x = 9$
Or $(3x + 7) = 0$
$\text{x}=\frac{-7}{3}$
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,
$(\text{x}-20)=69\times\frac{1}{\text{x}}$
$x(x - 20) = 69$
$x^2 - 20x - 69 = 0$
$x^2 - 23x + 3x - 69 = 0$
$x(x - 23) + 3(x - 23) = 0$
$(x - 23)(x + 3) = 0$
$(x - 23) = 0$
$x = 23$
$Or (x + 3) = 0$
$x = -3$
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 $p \neq q$, 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 $c=-2(p-q)$
$\Rightarrow D = {5(p + q)}^2 - 4(p - q)(-2(p - q))$
$\Rightarrow D = 25(p^2 + 2pq + q^2) + 8(p^2 - 2pq + q^2)$
$\Rightarrow D = 25p^2+ 50pq + 25q^2 + 8p^2 - 16pq + 8q^2$
$\Rightarrow D = 33p^2 + 34pq + 33q^2$
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$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6},$ $\text{x}\neq3,-5$
$\frac{1}{\text{x}-3}-\frac{1}{\text{x}+5}=\frac{1}{6}$
$\Rightarrow\frac{\text{x}+5-\text{x}+3}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow\frac{8}{(\text{x}-3)(\text{x}+5)}=\frac{1}{6}$
$\Rightarrow 48 = x^2 + 2x - 15$
$\Rightarrow x^2 + 2x - 15 - 48 = 0$
$\Rightarrow x^2 + 2x - 63 = 0$
$\Rightarrow x^2 + 9x - 7x - 63 = 0$
$\Rightarrow x(x + 9) - 7(x + 9) = 0$
$\Rightarrow (x - 7)(x + 9) = 0$
$\Rightarrow x = 7, -9$
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
$\text{x}+\text{x}^2=\frac{63}{4}$
$4(x + x^2) = 63$
$4x^2 + 4x - 63 = 0$
$4x^2 + 18x - 14x - 63 = 0$
$2x(2x + 9) - 7(2x + 9) = 0$
$(2x + 9)(2x - 7) = 0$
$(2x + 9) = 0$
$\text{x}=-\frac{9}{2}$
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 $x.$
$\Rightarrow $ The two integer will be x and $(x + 1)$
Product of two integers $\Rightarrow x(x + 1)$
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 $3x^2 + 13x + 14 = 0.$
AnswerGiven that the equation of $3x^2 + 13x + 14 = 0$
$3x^2 + 7x + 6x + 14 = 0$
$x(3x + 7) + 2(3x + 7) = 0$
$(3x + 7)(x + 2) = 0$
$(3x + 7) = 0$
$\text{x}=\frac{-7}{3}$
Or $(x + 2) = 0$
$x = -2$
Therefore, x = 2 is the solution of given equation.
Hence, proved.
View full question & answer→Question 1013 Marks
Solve for x.
$\text{x}+\frac{1}{\text{x}}=3,\text{x}\neq0$
Answer$\text{x}+\frac{1}{\text{x}}=3$
$\text{x}^2+1=3\text{x}$
$\text{x}^3-3\text{x}+1=0$
Here, $\text{a}=1,\text{b}=-3,\text{c}=1$
$\therefore\text{D}=\text{b}^2-4\text{ac}$
$=(-3)^2-4\times1\times1$
$=9-4=5$
$\because\text{D}>0$
$\therefore$ Roots are real
$\therefore\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$=\frac{-(-3)\pm\sqrt{5}}{2\times1}=\frac{(3)\pm\sqrt{5}}{2}$
$\therefore\text{x}=\frac{3+\sqrt{5}}{2},\frac{3-\sqrt{5}}{2}$
View full question & answer→Question 1023 Marks
Solve the following quadratic equations by factorization:
$9x^2 - 3x - 2 = 0$
AnswerWe have
$9x^2 - 3x - 2 = 0$
$\Rightarrow 9x^2 - 6x + 3x -2 = 0$
$\Rightarrow 3x(3x - 2) + 1(3x - 2) = 0$
$\Rightarrow (3x - 2)(3x + 1) = 0$
$\Rightarrow$ either $3x - 2 = 0 or 3x + 1 = 0$
$\Rightarrow 3x = 2 or 3x = -1$
$\Rightarrow\text{x}=\frac{2}{3}$ or $\text{x}=-\frac{1}{3}$
Thus, $\text{x}=\frac{2}{3}$ and $\text{x}=-\frac{1}{3}$ are two roots of the equation $9x^2 - 3x - 2 = 0$
View full question & answer→Question 1033 Marks
A train covers a distance of $90\ km$ at a uniform speed. Had the speed been 15km/hr more, it would have taken $30$ minutes less for the journey. Find the original speed of the train.
AnswerDistance to be covered = 90km
Let uniform-original speed = x km/h
Increased speed = (x + 15) km/hr
According to the condition,
$\frac{90}{\text{x}}-\frac{90}{\text{x}+15}=\frac{1}{2}$
$\big(30\text{ minutes}=\frac{1}{2}\text{hour}\big)$
$\Rightarrow\frac{90\text{x}+1350-90\text{x}}{\text{x}(\text{x}+15)}=\frac{1}{2}$
$\Rightarrow\frac{1350}{\text{x}^2+15\text{x}}=\frac{1}{2}$
$\Rightarrow\text{x}^2+15\text{x}=2700$
$\Rightarrow\text{x}^2+15\text{x}-2700=0$
$\begin{Bmatrix}\because-2700=60\times(-45)$
$15=60-45\end{Bmatrix} $
$\Rightarrow x^2 + 60x - 45x - 2700 = 0$
$\Rightarrow x(x + 60) - 45(x + 60) = 0$
$\Rightarrow (x + 60)(x - 45) =0$
Either $x + 60 = 0$, then $x = -60$ which is not possible being negative
or $x - 45 = 0,$ then $x = 45$
Original speed of the train = 45km/hr
View full question & answer→Question 1043 Marks
Write the set of values of ‘a’ for which the equation $x^2 + ax - 1 = 0$, has real roots.
Answer$x^2 + ax - 1 = 0$
Here $a = 1, b = a, c = -1$
$\Rightarrow D = b^2 - 4ac$
$\Rightarrow D = (a)^2 - 4 \times 1 \times (-1) = a^2 + 4$
Roots are real,
$\Rightarrow\text{D}\geq0$
$\Rightarrow\text{a}^2+4\geq0$
For all real values of a, the equation has real roots.
View full question & answer→Question 1053 Marks
Write the number of real roots of the equation $x^2 + 3 |x| + 2 = 0$
Answer$x^2 + 3 |x| + 2 = 0$
$x^2 + 3x + 2 = 0$
Here $x = 1, b = 3, c = 2$ $(\because|\text{x}|=\text{x})$
$D = b^2 - 4ac$
$D = (3)^2 - 4 \times 1 \times 2$
$D = 9 - 8$
$D = 1$
$\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\text{x}=\frac{-3\pm\sqrt{1}}{2\times1}$
$\text{x}=\frac{-3\pm{1}}{2}$
$\text{x}_1=\frac{-3+1}{2}$
$\text{x}_1=\frac{-2}{2}$
$\text{x}_1=-1$
and $\text{x}_2=\frac{-3-1}{2}$
$\text{x}_2=\frac{-4}{2}$
$\text{x}_2=-2$
$\therefore$ Real roots are -1, -2
View full question & answer→Question 1063 Marks
Find the value of k for which the following equation have real root:
$x^2 - 4kx + k = 0$
Answer$x^2 - 4kx + k = 0$
Here, comparing with $ax^2 + kx + c = 0$
$a = 1, b = -4k, c = k$
$D = b^2 - 4ac$
$D = (-4k)^2- 4 \times 1 \times k$
$D = 16k^2 - 4k$
$\because$ Roots are real and equal
$\therefore D = 0$
$\therefore 16k^2 - 4k = 0$
$\Rightarrow 4k(4k - 1) = 0$
$\Rightarrow k(4k - 1) = 0$
Either $k = 0$
$or 4k - 1 = 0$
$4k = 1$
$\therefore\text{k}=\frac{1}{4}$
Hence k = 0, $\frac{1}{4}$
View full question & answer→Question 1073 Marks
If $1+\sqrt{2}$ is a root of a quadratic equation will rational coefficients, write its other root.
AnswerGiven that $(1+\sqrt{2})$ is a root of the quadratic equation with rational coefficients.
Then find the other root.
As we know that if $(1+\sqrt{2})$ is a root of the quadratic equation with rational.
coefficients then other roots be $(1+\sqrt{2})$
Hence, the require root of the quadratic equation be $(1+\sqrt{2})$
View full question & answer→Question 1083 Marks
A car moves a distance of 2592km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.
AnswerDistance = 2592km
Let the speed of the car = x km/hr
and time taken $=\frac{\text{x}}{2}\text{hr}$
We have, Distance = Speed × Time
$\Rightarrow2592=\text{x}\times\frac{\text{x}}{2}$
$\Rightarrow2592=\frac{\text{x}^2}{2}$
$\Rightarrow\text{x}^2=2592\times2$
$\Rightarrow\text{x}=\sqrt{5184}$
$\Rightarrow\text{x}=72\text{km/hr}$
and thus time taken $=\frac{\text{x}}{2}\text{h}=\frac{72}{2}=36\text{ hr}$
View full question & answer→Question 1093 Marks
Find two consecutive natural numbers whose product is $20.$
AnswerLet the two consecutive natural numbers be $'x'$ and '$x + 1'$
Given that product of the natural numbers is $20$
Hence, $x(x + 1) = 20$
$\Rightarrow x^2 + x = 20$
$\Rightarrow x^2 + x - 20 = 0$
$\Rightarrow x^2 + 5x - 4x - 20 = 0$
$\Rightarrow x(x + 5) - 4(x + 5) = 0$
$\Rightarrow x = -5 or x = 4$
Considering positive value of x as $\text{x}\in\text{N}$
For $r = 4, x + 1 = 4 + 1 = 5$
The two consecutive natural numbers are $4$ as $5$
View full question & answer→Question 1103 Marks
Write a quadratic polynomial, sum of whose zeros is $2\sqrt{3}$ and their product is $2.$
AnswerAs we know that the quadratic
polynomial $f(x) = k[x^2 - $(sum of their roots)x + (product of their roots)]
According to question,
(sum of their roots) $=2\sqrt{3}$
And (product of their roots) = 2
Thus Putting the value in above,
$\text{f}(\text{x})=\text{k}\big[\text{x}^2-2\sqrt{3}\text{x}+2\big]$ where k is real number.
Therefore, the quadratic polynomial be
$\text{f}(\text{x})=\text{k}\big[\text{x}^2-2\sqrt{3}\text{x}+2\big]$
View full question & answer→Question 1113 Marks
The difference of the squares of two positive integers is $180$. The square of the smaller number is $8$ times the larger, find the numbers.
AnswerLet the larger number be x
Then according to the question,
Square of the smaller number $= 8x$, then
$\Rightarrow x^2 - 8x = 180$
$\Rightarrow x^2 - 8x - 180 = 0$
$\Rightarrow x^2- 18x + 10x - 180 = 0$
$\Rightarrow x(x - 18) + 10(x - 18) = 0$
$\Rightarrow (x + 10)(x - 18) = 0$
$\Rightarrow x + 10 = 0 or x - 18 = 0$
$\Rightarrow x = -10 or x = 18$
Since, x being a positive integer so, x cannot be negative,
Therefore, larger number $= 18$
Then the smaller number $=\sqrt{8\times18}=12$
Thus, the two positive number are $12$ and $18$
View full question & answer→Question 1123 Marks
There are three consecutive integers such that the square of the first increased by the product of the other two gives $154$. What are the integers.
AnswerLet first integer $= x$
Then second integer $= x + 1$, and third integer $= x + 2$
According to the condition,
$\Rightarrow x^2 + (x + 1)(x + 2) = 154$
$\Rightarrow x^2 + x^2 + 3x + 2 = 154$
$\Rightarrow 2x^2 + 3x + 2 - 154 = 0$
$\Rightarrow 2x^2 - 16x + 19x - 152 = 0$
$\Rightarrow 2x(x - 8) + 19(x - 8) = 0$
$\Rightarrow (x - 8)(2x + 9) = 0$
Either x - 8 = 0, then x = 8
or $2x + 19 = 0,$ then $2x = -19$
$\Rightarrow\text{x}=\frac{-19}{2}$ But it is not an integer.
First number $= 8$
Second number $= 8 + 1 = 9$ and third number $= 8 + 2 = 10$
View full question & answer→Question 1133 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$\sqrt{2}\text{x}^2+7\text{x}+5\sqrt{2}=0$
Answer$\sqrt{2}\text{x}^2+7\text{x}+5\sqrt{2}=0$
The given equation is in the form of $\text{ax}^2+\text{bx}+\text{c}=0$
Here, $\text{a}=\sqrt{2},\text{b}=7,\text{c}=5\sqrt{2}$
The discriminant $\text{D}=\text{b}^2-4\text{ac}$
$=(7)^2-4\times\sqrt{2}\times5\sqrt{2}$
$=49-40$
$=9>0$
As 0 = 0, the given equation has roots real given by
$\Rightarrow\alpha=-\frac{\text{b}+\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-7+\sqrt{9}}{2\times\sqrt{2}}=\frac{-7+3}{2\sqrt{2}}$
$=\frac{-\text{4}}{2\sqrt{2}}=-\sqrt{2}$
$\Rightarrow\beta=-\frac{\text{b}-\sqrt{\text{D}}}{2\text{a}}$
$=\frac{-7-\sqrt{9}}{2\sqrt{2}}=\frac{-7-3}{2\sqrt{2}}$
$=\frac{-10}{2\sqrt{2}}=\frac{-5}{\sqrt{2}}$
$\therefore$ The roots of the given equation are $-\sqrt{2},\frac{-5}{\sqrt{2}}$
View full question & answer→Question 1143 Marks
Determine the nature of the root of following quadratic equation:
$(b + c)x^2 - (a + b + c)x + a = 0$
AnswerThe given quadric equation is $(b + c)x^2 - (a + b + c)x + a = 0$
Here, $a = (b + c), b = -(a + b + c)$ and $c = a$
As we know that $D = b^2 - 4ac$
Putting the value of $a = (b + c), b = -(a + b + c)$ and $c = a$
$= (-(a + b + c))^2 - 4 \times (b + c) \times a$
$= (a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - 4ab - 4ca$
$= a^2 + b^2 + c^2 - 2ab + 2bc - 2ca$
Since, $D > 0$
Therefore, root of the given equation are real and unequal.
View full question & answer→Question 1153 Marks
In the following, determine whether the given values are solution of the given equation or not:
$\text{x}^2-3\sqrt{3}\text{x}+6=0,$ $\text{x}=\sqrt{3},\text{x}=-2\sqrt{3}$
AnswerWe have been given that,
$\text{x}^2-3\sqrt{3}\text{x}+6=0,$ $\text{x}=\sqrt{3},\text{x}=-2\sqrt{3}$
Now, if $\text{x}=\sqrt{3}$ is a solution then it should satisfy the equation so, substituting $\text{x}=\sqrt{3}$ in the equation, we get
$\text{x}^2-3\sqrt{3}\text{x}+6=(\sqrt{3})^2-3\sqrt{3}(\sqrt{3})+6$
$=3-9+6$
$=0$
$\text{x}=-2\sqrt{3}$
$(-2\sqrt{3})^2-3\sqrt{3}(-2\sqrt{3})+6$
$4\times3+6\times3+6$
$=12+18+6$
$=36$
Which in not equal to zero.
Hence, $\text{x}=-2\sqrt{3}$ is a solution of the quadratic equation.
Therefore, from the above results we find out that $\text{x}=\sqrt{3}$ and $\text{x}=-2\sqrt{3}$ is not solutions of the given quadratic equation.
View full question & answer→Question 1163 Marks
Determine the set of values of k for which the given following quadratic has real root:
$2x^2 + kx + 2 = 0$
AnswerThe given quadric equation is $2 x^2+k x+2=0$, and roots are real.
Then find the value of $k$.
Here, $a=2, b=k$ and $c =2$
As we know that of $a =2, b= k$ and $c =2$
$= (k)^2 - 4 \times 2 \times 2$
$= k^2 - 16$
The given equation will have real roots, if $\text{D}\geq0$
$\text{k}^2-16\geq0$
$\text{k}^2\geq16$
$\text{k}\geq\sqrt{16}$ or $\text{k}\leq-\sqrt{16}$
$\text{k}\leq-4$ or $\text{k}\geq4$
Therefore, the value of $\text{k}\leq-4$ or $\text{k}\geq4$
View full question & answer→Question 1173 Marks
Find the value of k for which the root are real and equal in the following equations:
$3x^2 - 5x + 2k = 0$
AnswerThe given equation is $3x^2 - 5x + 2k = 0$
This equation is in the from of $ax^2 + bx + c = 0$
Here, $a = 3, b = -5$ and $c = 2k$
Given that, the equation has real and equal roots
i.e.,$ D = b^2 - 4ac = 0$
$\Rightarrow (-5)^2 - 4 \times 3 \times (2k) = 0$
$\Rightarrow 25 = 24k$
$\Rightarrow\text{k}=\frac{25}{24}$
$\therefore$ The value of $\text{k}=\frac{25}{24}$
View full question & answer→Question 1183 Marks
Find the roots of the quadratic equation:
$\sqrt{2}\text{x}^2+7\text{x}+5\sqrt{2}=0$
Answer$\sqrt{2}\text{x}^2+7\text{x}+5\sqrt{2}=0$
$\Rightarrow\sqrt{2}\text{x}^2+5\text{x}+2\text{x}+5\sqrt{2}=0$
$\Rightarrow\text{x}\big(\sqrt{2}\text{x}+5\big)+\sqrt{2}\big(\sqrt{2}\text{x}+5\big)=0$
$\Rightarrow\big(\sqrt{2}\text{x}+5\big)\big(\text{x}+\sqrt{2}\big)=0$
Either $\sqrt{2}\text{x}+5=0,$ then $\text{x}=\frac{-5}{\sqrt{2}}$
or $\text{x}+\sqrt{2}=0,$ then $\text{x}=-\sqrt{2}$
$\therefore$ Roots are $\frac{-5}{\sqrt{2}},-\sqrt{2}$
View full question & answer→Question 1193 Marks
The difference of two numbers is $4.$ If the difference of their reciprocal is $\frac{4}{21},$ find the numbers.
AnswerLet the two numbers be x and $x - 4$
Given that the difference of two numbers is $4$
By the given hypothesis, we have $\frac{1}{\text{x}-4}-\frac{1}{\text{x}}=\frac{4}{21}$
$\Rightarrow\frac{\text{x}-\text{x}+4}{\text{x}(\text{x}-4)}=\frac{4}{21}$
$\Rightarrow 84 = 4x(x - 4)$
$\Rightarrow x^2 - 4x - 21 = 0$
$\Rightarrow x^2 - 7x + 3x - 21 = 0$
$\Rightarrow x(x - 7) + 3(x - 7) = 0$
$\Rightarrow (x - 7)(x + 3) = 0$
$\Rightarrow x = 7 or x = -3$ and
If $x = -3, x - 4 = -3 - 4 = -7$
Hence, required numbers are $3, 7$ and $-3, -7x$
View full question & answer→Question 1203 Marks
A train travels $360\ km$ at a uniform speed. If the speed had been 5km/hr more, it would have taken $1$ hour less for the same journey. Form the quadratic equation to find the speed of the train.
AnswerTotal distance = 360km
Let the uniform speed of the train = x km/hr
Time taken $=\frac{360}{\text{x}}$
In second case speed = (x + 5) km/hr
$\therefore$ Time taken $=\frac{360}{\text{x}+5}\text{h}$
$\therefore\frac{360}{\text{x}}-\frac{360}{\text{x}+5}=1$
$\Rightarrow 360x + 1800 - 360x = x(x + 5)$
$\Rightarrow 1800 = x^2 + 5x$
$\Rightarrow x^2 + 5x - 1800 = 0$
$\therefore$ Required quadratic equation will be
$x^2 + 5x - 1800 = 0$
View full question & answer→Question 1213 Marks
Solve the following quadratic equations by factorization:
$16\text{x}-\frac{10}{\text{x}}=27$
Answer$16\text{x}-\frac{10}{\text{x}}=27$
$16x^2 - 10 = 27x$
$16x^2 - 27x - 10 = 0$
$16x^2 - 32x + 5x - 10 = 0$
$16x(x - 2) + 5 (x - 2) = 0$
$(16x + 5)(x - 2) = 0$
$16x + 5 = 0 or x - 2 = 0$
$\Rightarrow\text{x}=-\frac{5}{16}$ or $x = 2$
Hence, the factors are 2 and $-\frac{5}{16}$
View full question & answer→Question 1223 Marks
Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
$x^2 - 4ax + 4a^2 - b^2 = 0$
Answer$\text{x}^2-4\text{ax}+4\text{a}^2-\text{b}^2=0$
$\Rightarrow\text{x}^2-2\cdot2\text{a}\cdot\text{x}+(2\text{a})^2=\text{b}^2$
$\Rightarrow(\text{x}-2\text{a})^2=(\pm\text{b})^2$
$\Rightarrow\text{x}-2\text{a}=\pm\text{b}$
$\therefore\text{x}=2\text{a}\pm\text{b}$
$\therefore\text{x}=2\text{a}+\text{b}$
and $\text{x}=2\text{a}-\text{b}$
Hence roots are $2\text{a}+\text{b},2\text{a}-\text{b}$
View full question & answer→Question 1233 Marks
Solve the following quadratic equation by factorization:
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
AnswerWe have
$a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$[-1 \times a^2b^2 = -a^2b^2$
$\Rightarrow -a^2b^2 = -a^2 \times b^2 = -a^2 \times b^2]$
$\Rightarrow a^2b^2x^2 + b^2x - a^2x - 1 = 0$
$\Rightarrow b^2x(a^2x + 1) - 1(a^2x + 1) = 0$
$\Rightarrow (a^2x + 1)(b^2x - 1) = 0$
$\Rightarrow a^2x + 1 = 0 or b^2x - 1 = 0$
$\Rightarrow\text{x}= -\frac{1}{\text{a}^2}$ and $\text{x}= \frac{1}{\text{b}^2}$ are the two root of the given equation.
View full question & answer→Question 1243 Marks
In the following, determine the set of values of k for which the given quadratic equation has real root:
$2x^2 + x + k = 0$
AnswerThe given equation is $2x^2 + x + k = 0$, and roots are real.
Then find the value of $k$.
Here, $a = 2, b = 1, c = k$
As we know that$ D = b^2 - 4ac$
Putting the value of $a = 2, b = 1, c = k$
$D = 1 - 8k$
The given equation will have real roots, if $\text{D}\geq0$
$\text{D}=1-8\text{k}\geq0$
$\Rightarrow8\text{k}\leq1$
$\Rightarrow\text{k}\leq\frac{1}{8}$
Therefore, the value of $\text{k}\leq\frac{1}{8}$
View full question & answer→Question 1253 Marks
Determine the nature of the root of following quadratic equation:
$2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$
Answer$2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$
Here, $a = 2(a^2 + b^2), b = 2(a + b), c = 1$
$\therefore$ Discriminant $(D) = b^2 - 4ac$
$= {2(a + b)}^2 - 4 \times 2(a^2 + b^2) \times 1$
$= 4(a^2 + b^2 + 2ab) - 8(a^2 + b^2)$
$= 4a^2 + 4b^2 + 8ab - 8a^2 - 8b^2$
$= 8ab - 4a^2 - 4b^2$
$= -(4a^2 + 4b^2 - 8ab)$
$= -4(a^2 + b^2 - 2ab)$
$= -4(a - b)^2$^
$\because$
$D < 0$
$\therefore$ Roots are not real.
View full question & answer→Question 1263 Marks
The sum of the squares of two consecutive odd numbers is $394$. Find the numbers.
AnswerLet the consecutive odd positive integers are $2x - 1$ and $2x + 1$
Given that the sum of the squares is $394$
$\Rightarrow (2x - 1)^2 + (2x + 1)^2 = 394$
$\Rightarrow 4x^2 + 1 - 4x + 4x^2 + 1 + 4x = 394$
$\Rightarrow 8x^2 + 2 = 394$
$\Rightarrow 8x^2 = 394 - 2$
$\Rightarrow 8x^2 = 392$
$\Rightarrow\text{x}^2=\frac{392}{8}$
$x = 7$
$As x = 7, 2x - 1 = 2 \times 7 - 1 = 14 - 1 = 13$
$\Rightarrow 2x + 1 = 2 \times 7 + 1 = 14 + 1 = 15$
The two consecutive odd positive numbers are $11$ and $13$.
View full question & answer→Question 1273 Marks
Find the roots of the following quadratic equation (if they exist) by the method of completing the square.
$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
Answer$\sqrt{2}\text{x}^2-3\text{x}-2\sqrt{2}=0$
$\Rightarrow\text{x}^2-\frac{3}{\sqrt{2}}\text{x}-2=0$
$($Dividng by $\sqrt{2})$
$\Rightarrow(\text{x}^2)-2\times\frac{3}{2\sqrt{2}}\text{x}+\Big(\frac{3}{2\sqrt{2}}\Big)^2-\frac{25}{8}=0$
$\Rightarrow\Big(\text{x}-\frac{3}{2\sqrt{2}}\Big)^2=\Big(\pm\frac{5}{2\sqrt{2}}\Big)^2$
$\begin{cases}\because-2=\frac{9}{8}-\frac{25}{8}\\=\frac{3}{2\sqrt{2}}-\frac{25}{8}\end{cases}$
$\therefore\text{x}-\frac{3}{2\sqrt{2}}=\pm\frac{5}{2\sqrt{2}}$
$\Rightarrow\text{x}=\frac{3}{2\sqrt{2}}\pm\frac{5}{2\sqrt{2}}$
$=\frac{3}{2\sqrt{2}}\pm\frac{5}{2\sqrt{2}}$
$=\frac{8}{2\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$
and $\text{x}=\frac{3}{2\sqrt{2}}-\frac{5}{2\sqrt{2}}$
$=\frac{-2}{2\sqrt{2}}=\frac{-1}{\sqrt{2}}$
Roots are $=2\sqrt{2}$ and $\frac{-1}{\sqrt{2}}$
View full question & answer→Question 1283 Marks
In the following, determine whether the given quadratic equation have real root and if so, find the root:
$x^2 - 2x + 1 = 0$
Answer$x^2 - 2x + 1 = 0$
The given equation is in the form of $ax^2 + bx + c = 0$
Here $a = 1, b = -2$ and $c = 1$
The discriminant $D = b^2 - 4ac$
$\Rightarrow (-2)^2 - 4 \times 1 \times 1 = 0$
As Q = 0, the given equation has real and equal roots
$\Rightarrow\alpha=-\frac{\text{b}+\sqrt{\text{D}}}{2\text{a}},$
$\beta=-\frac{\text{b}-\sqrt{\text{D}}}{2\text{a}}$
i.e., $\alpha$ and $\beta=-\frac{\text{b}}{2\text{a}}\ [\because0=0]$
$\Rightarrow\alpha$ and $\beta=-\frac{\text{b}}{2\text{a}}$
$=-\frac{(-2)}{2\times2}=\frac{2}{2}=1$
The roots of the given equation $\alpha$ and $\beta$ is 1
View full question & answer→Question 1293 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 $\frac{-\sqrt{3}}{2},-2\sqrt{3}$
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