# CAT Quadratic Equation Questions PDF [Most Important]

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Quadratic Equations is an important topic in the CAT Quant (Algebra) section. If you find these questions a bit tough, make sure you solve more CAT Quadratic Equation questions. Learn all the important formulas and tricks on how to answer questions on Quadratic Equations. You can check out these Quadratic Equation questions from the Practice a good number of sums in the CAT Quadratic Equation so that you can answer these questions with ease in the exam. In this post, we will look into some important Quadratic Equation Questions in the CAT quants. These are a good source of practice for CAT 2022 preparation; If you want to practice these questions, you can download this Important CAT Quadratic Equation Questions and answers PDF along with the video solutions below, which are completely Free.

Question 1:Â A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f (x) at x = 10?

a)Â -119

b)Â -159

c)Â -110

d)Â -180

e)Â -105

Solution:

Let the function be $ax^2 + bx + c$.

We know that x=0 value is 1 so c=1.

So equation is $ax^2 + bx + 1$.

Now max value is 3 at x = 1.

So after substituting we get a + b = 2.

If f(x) attains a maximum at ‘a’ then the differential of f(x) at x=a, that is, f'(a)=0.

So in this question f'(1)=0

=> 2*(1)*a+b = 0

=> 2a+b = 0.

Solving the equations we get a=-2 and b=4.

$-2x^2 + 4x + 1$ is the equation and on substituting x=10, we get -159.

Question 2:Â If the roots of the equation $x^3 – ax^2 + bx – c = 0$ are three consecutive integers, then what is the smallest possible value of b?
[CAT 2008]

a)Â $\frac{-1}{\sqrt 3}$

b)Â $-1$

c)Â $0$

d)Â $1$

e)Â $\frac{1}{\sqrt 3}$

Solution:

b = sum of the roots taken 2 at a time.
Let the roots be n-1, n and n+1.
Therefore, $b = (n-1)n + n(n+1) + (n+1)(n-1) = n^2 – n + n^2 + n + n^2 – 1$
$b = 3n^2 – 1$. The smallest value is -1.

Question 3:Â Let p and q be the roots of the quadratic equation $x^2 – (\alpha – 2) x – \alpha -1= 0$ . What is the minimum possible value of $p^2 + q^2$?

a)Â 0

b)Â 3

c)Â 4

d)Â 5

Solution:

Let $\alpha$ be equal to k.

=> f(x) = $x^2-(k-2)x-(k+1) = 0$

p and q are the roots

=> p+q = k-2 and pq = -1-k

We know that $(p+q)^2 = p^2 + q^2 + 2pq$

=> $(k-2)^2 = p^2 + q^2 + 2(-1-k)$

=> $p^2 + q^2 = k^2 + 4 – 4k + 2 + 2k$

=> $p^2 + q^2 = k^2 – 2k + 6$

This is in the form of a quadratic equation.

The coefficient of $k^2$ is positive. Therefore this equation has a minimum value.

We know that the minimum value occurs at x = $\frac{-b}{2a}$

Here a = 1, b = -2 and c = 6

=> Minimum value occurs at k = $\frac{2}{2}$ = 1

If we substitute k = 1 in $k^2-2k+6$, we get 1 -2 + 6 = 5.

Hence 5 is the minimum value that $p^2+q^2$ can attain.

Question 4:Â Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the original quadratic equation?

a)Â (6, 1)

b)Â (-3, -4)

c)Â (4, 3)

d)Â (-4, -3)

Solution:

We know that quadratic equation can be written as $x^2$-(sum of roots)*x+(product of the roots)=0.
Ujakar ended up with the roots (4, 3) so the equation is $x^2$-(7)*x+(12)=0 where the constant term is wrong.

Keshab got the roots as (3, 2) so the equation is $x^2$-(5)*x+(6)=0 where the coefficient of x is wrong .

So the correct equation is $x^2$-(7)*x+(6)=0. The roots of above equations are (6,1).

Question 5:Â If the roots $x_1$ and $x_2$ are the roots of the quadratic equation $x^2 -2x+c=0$ also satisfy the equation $7x_2 – 4x_1 = 47$, then which of the following is true?

a)Â c = -15

b)Â $x_1 = -5$ and $x_2=3$

c)Â $x_1 = 4.5$ and $x_2 = -2.5$

d)Â None of these

Solution:

$x_1 + x_2 = 2$
and $7x_2 – 4x_1 = 47$
So $x_1 = -3$ and $x_2 = 5$
And $c = x_1 \times x_2 = -15$

Question 6:Â If $x+1=x^{2}$ and $x>0$, then $2x^{4}$Â  is

a)Â $6+4\sqrt{5}$

b)Â $3+3\sqrt{5}$

c)Â $5+3\sqrt{5}$

d)Â $7+3\sqrt{5}$

Solution:

We know that $x^2 – x – 1=0$
Therefore $x^4 = (x+1)^2 = x^2+2x+1 = x+1 + 2x+1 = 3x+2$
Therefore, $2x^4 = 6x+4$

We know that $x>0$ therefore, we can calculate the value of $x$ to be $\frac{1+\sqrt{5}}{2}$
Hence, $2x^4 = 6x+4 = 3+3\sqrt{5}+4 = 3\sqrt{5}+7$

Question 7:Â If $U^{2}+(U-2V-1)^{2}$= âˆ’$4V(U+V)$ , then what is the value of $U+3V$ ?

a)Â $0$

b)Â $\dfrac{1}{2}$

c)Â $\dfrac{-1}{4}$

d)Â $\dfrac{1}{4}$

Solution:

Given that $U^{2}+(U-2V-1)^{2}$= âˆ’$4V(U+V)$

$\Rightarrow$ $U^{2}+(U-2V-1)(U-2V-1)$= âˆ’$4V(U+V)$

$\Rightarrow$ $U^{2}+(U^2-2UV-U-2UV+4V^2+2V-U+2V+1)$ =Â âˆ’$4V(U+V)$

$\Rightarrow$ $U^{2}+(U^2-4UV-2U+4V^2+4V+1)$ =Â âˆ’$4V(U+V)$

$\Rightarrow$ $2U^2-4UV-2U+4V^2+4V+1=âˆ’4UV-4V^2$

$\Rightarrow$ $2U^2-2U+8V^2+4V+1=0$

$\Rightarrow$ $2[U^2-U+\dfrac{1}{4}]+8[V^2+\dfrac{V}{2}+\dfrac{1}{16}]=0$

$\Rightarrow$ $2(U-\dfrac{1}{2})^2+8(V+\dfrac{1}{4})^2=0$

Sum of two square terms is zero i.e. individual square term is equal to zero.

$U-\dfrac{1}{2}$ = 0 and $V+\dfrac{1}{4}$ = 0

U = $\dfrac{1}{2}$ andÂ V = $-\dfrac{1}{4}$

Therefore,Â $U+3V$ =Â $\dfrac{1}{2}$+$\dfrac{-1*3}{4}$ =Â $\dfrac{-1}{4}$. Hence, option C is the correct answer.

Question 8:Â If a and b are integers such that $2x^2âˆ’ax+2>0$ and $x^2âˆ’bx+8â‰¥0$ for all real numbers $x$, then the largest possible value of $2aâˆ’6b$ is

Solution:

Let f(x) = $2x^2âˆ’ax+2$. We can see that f(x) is a quadratic function.

For, f(x) > 0, Discriminant (D) < 0

$\Rightarrow$ $(-a)^2-4*2*2<0$

$\Rightarrow$ (a-4)(a+4)<0

$\Rightarrow$ a $\epsilon$ (-4, 4)

Therefore, integer values that ‘a’ can take = {-3, -2, -1, 0, 1, 2, 3}

Let g(x) = $x^2âˆ’bx+8$. We can see that g(x) is also a quadratic function.

For, g(x)â‰¥0, Discriminant (D) $\leq$ 0

$\Rightarrow$ $(-b)^2-4*8*1<0$

$\Rightarrow$ $(b-\sqrt{32})(b+\sqrt{32})<0$

$\Rightarrow$ bÂ $\epsilon$ (-$\sqrt{32}$, $\sqrt{32}$)

Therefore, integer values that ‘b’ can take = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}

We have to find out theÂ largest possible value of $2aâˆ’6b$. The largest possible value will occur when ‘a’ is maximum and ‘b’ is minimum.

a$_{max}$ = 3,Â b$_{min}$ = -5

Therefore, theÂ largest possible value of $2aâˆ’6b$ = 2*3 – 6*(-5) = 36.

Question 9:Â Let A be a real number. Then the roots of the equation $x^2 – 4x – log_{2}{A} = 0$ are real and distinct if and only if

a)Â $A > \frac{1}{16}$

b)Â $A < \frac{1}{16}$

c)Â $A < \frac{1}{8}$

d)Â $A > \frac{1}{8}$

Solution:

The roots ofÂ $x^2 – 4x – log_{2}{A} = 0$ will be real and distinct if and only if the discriminant is greater than zero

16+4*$log_{2}{A}$ > 0

$log_{2}{A}$ > -4

A> 1/16

Question 10:Â The quadratic equation $x^2 + bx + c = 0$ has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of $b^2 + c$?

a)Â 3721

b)Â 361

c)Â 427

d)Â 549

Solution:

Given,

The quadratic equation $x^2 + bx + c = 0$ has two roots 4a and 3a

7a=-b

12$a^2$ = c

We have to find the value ofÂ  $b^2 + c$ = 49$a^2$+ 12$a^2$=61$a^2$

Now lets verify the options

61$a^2$ = 3721 ==> a= 7.8 which is not an integer

61$a^2$ = 361 ==> a= 2.42 which is not an integer

61$a^2$ = 427 ==> a= 2.64 which is not an integer

61$a^2$ = 3721 ==> a= 3 which is an integer

Question 11:Â The product of the distinct roots of $\mid x^2 – x – 6 \mid = x + 2$ is

a)Â âˆ’16

b)Â -4

c)Â -24

d)Â -8

Solution:

We have,Â $\mid x^2 – x – 6 \mid = x + 2$

=> |(x-3)(x+2)|=x+2

For x<-2, (3-x)(-x-2)=x+2

=> x-3=1Â  Â =>x=4 (Rejected as x<-2)

For -2$\le\$x<3,Â (3-x)(x+2)=x+2Â  Â  =>x=2,-2

For x$\ge\$3,Â (x-3)(x+2)=x+2Â  Â =>x=4

Hence the product =4*-2*2=-16

Question 12:Â The number of solutions to the equation $\mid x \mid (6x^2 + 1) = 5x^2$ is

Solution:

For x <0, -x($6x^2+1$) =Â $5x^2$

=>Â ($6x^2+1$) = -5x

=>Â ($6x^2 + 5x+ 1$) = 0

=>($6x^2 + 3x+2x+ 1$) = 0

=> (3x+1)(2x+1)=0Â  Â  =>x=$\ -\frac{\ 1}{3}$Â  or x=$\ -\frac{\ 1}{2}$

For x=0, LHS=RHS=0Â  Â  (Hence, 1 solution)

For x >0, x($6x^2+1$) = $5x^2$

=> ($6x^2 – 5x+ 1$) = 0

=>(3x-1)(2x-1)=0Â  Â  =>x=$\ \frac{\ 1}{3}$Â  Â orÂ  Â x=$\ \frac{\ 1}{2}$

Hence, the total number of solutions = 5

Question 13:Â How many disticnt positive integer-valued solutions exist to the equation $(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$ ?

a)Â 8

b)Â 4

c)Â 2

d)Â 6

Solution:

$(x^{2}-7x+11)^{(x^{2}-13x+42)}=1$

ifÂ $(x^{2}-13x+42)$=0 orÂ $(x^{2}-7x+11)$=1 orÂ $(x^{2}-7x+11)$=-1 andÂ $(x^{2}-13x+42)$ is even number

For x=6,7 the value $(x^{2}-13x+42)$=0

$(x^{2}-7x+11)$=1 for x=5,2.

$(x^{2}-7x+11)$=-1 for x=3,4 and for X=3 or 4,Â $(x^{2}-13x+42)$ is even number.

.’. {2,3,4,5,6,7} is the solution set of x.

.’. x can take six values.

Question 14:Â The number of distinct real roots of the equation $(x+\frac{1}{x})^{2}-3(x+\frac{1}{x})+2=0$ equals

Solution:

Let $a=x+\frac{1}{x}$
So, the given equation is $a^2-3a+2=0$
So, $a$ can be either 2 or 1.

If $a=1$, $x+\frac{1}{x}=1$ and it has no real roots.
If $a=2$, $x+\frac{1}{x}=2$ and it has exactly one real root which is $x=1$

So, the total number of distinct real roots of the given equation is 1

Question 15:Â Let m and n be positive integers, If $x^{2}+mx+2n=0$ and $x^{2}+2nx+m=0$ have realÂ roots, then the smallest possible value of $m+n$ is

a)Â 7

b)Â 6

c)Â 8

d)Â 5

So, $m^2-8n\ge0\ \&\ 4n^2-4m\ge0$
=> $m^2\ge8n\ \&\ n^2\ge m$