CAT Quadratic Equations Questions & Solutions

A sample of real CAT Quadratic Equations past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 48 Quadratic Equations questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   A value of $c$ for which the minimum value of $f(x)=x^{2}-4cx+8c$ is greater than the maximum value of $g(x)=-x^{2}+3cx-2c$, is

   • $2$

   • $\dfrac{1}{2}$

   • $-\dfrac{1}{2}$

   • $-2$

   Show solution

   First function  $f\left(x\right)=x^2-4cx+8$

   For this function a>0 , so minimum value will occur at  $x=-\dfrac{b}{2a}=-\left(-\dfrac{4c}{2}\right)=2c$

   So, the minimum value of the function is =  $2c^2-4c\left(2c\right)+8c=-4c^2+8c$

   Second function  $g(x)=-x^{2}+3cx-2c$

   For this function a<0 , so maximum value will occur at  $x=-\dfrac{b}{2a}=-\dfrac{\left(-3c\right)}{2}=\dfrac{3c}{2}$

   So, the maximum value of the function is =  $-\left(\dfrac{3c}{2}\right)^2+3c\left(\dfrac{3c}{2}\right)-2c=\dfrac{9c^2}{4}-2c$

   So, as per the given condition,

   \dfrac{9c^2}{4}-2c<-4c^2+8c

   or,  \dfrac{9c^2}{4}+4c^2<8c+2c

   or,  \dfrac{25c^2}{4}<10c

   or,  \dfrac{5c^2}{4}<2c

   or,  5c^2<8c

   or,  5c^2-8c<0

   or,  c\left(c-\dfrac{8}{5}\right)<0

   or,  0 < c < \dfrac{8}{5}

   So, the value of $c$ which lies in this range is  $\dfrac{1}{2}$

2. Q2.CAT 2025

   The equations $3x^{2}-5x+p=0$ and $2x^{2}-2x+q=0$ have one common root. The sum of the other roots of this equations is

   • $\frac{8}{3}-p+\frac{3}{2}q$

   • $\frac{2}{3}-p+\frac{3}{2}q$

   • $\frac{8}{3}+p+\frac{1}{3}q$

   • $\frac{2}{3}-2p+\frac{2}{3}q$

   Show solution

   Let's assume that the common root is r. 

   The sum of the roots of the first equation is 5/3 and that of the second equation is 1.

   We want the sum of the other two roots:

   $\text{Sum} = \left(\frac{5}{3}-r\right) + (1-r) = \frac{8}{3}-2r$
   

   We now need to express r in terms of p and q.

   Since r is a common root, it satisfies:

   $3r^2 - 5r + p = 0 \quad (1)$

   $2r^2 - 2r + q = 0 \quad (2)$
   

   Eliminate $r^2$ .

   Multiply (2) by 3:

   $6r^2 - 6r + 3q = 0$
   

   Multiply (1) by 2:

   $6r^2 - 10r + 2p = 0$
   

   Subtract:

   $(6r^2 - 6r + 3q) - (6r^2 - 10r + 2p) = 0$

   $4r + 3q - 2p = 0$

    r = \frac{2p - 3q}{4}
   

   Now substitute into $\frac{8}{3} - 2r$ :

   $\frac{8}{3} - 2\left(\frac{2p - 3q}{4}\right)$

   $= \frac{8}{3} - \frac{2p - 3q}{2}$

   $= \frac{8}{3} - p + \frac{3}{2}q$
   

3. Q3.CAT 2024

   lf the equations $x^{2}+mx+9=0, x^{2}+nx+17=0$ and $x^{2}+(m+n)x+35=0$ have a common negative root, then the value of $(2m+3n)$ is

   Answer: 38

   Show solution

   When given more than one equations, stating the fact that there is a common root, 
   We need to equate the two equations to get discernible values for $x$

   Here, we are given three equations with the values of $m$ , $n$

   $x^2+mx+9=x^2+\left(m+n\right)x+35$
   $mx+9=mx+nx+35$
   $nx=-26$

   Similarly, we can do it for the other equation as well, 
   $x^2+nx+17=x^2+\left(m+n\right)x+35$
   $mx=-18$

   Substituting the value of either $mx$ or $nx$ in the original equations, we get 
   $x^2-18+9=0$
   $x^2=9$
   $x=\pm\ 3$
   Since we are given that the root is negative,  $x=-3$

   $n=-\dfrac{26}{-3}$
   $m=-\dfrac{18}{-3}$

   $3n=26$
   $2m=12$

   $2m+3n=38$

4. Q4.CAT 2023

   The sum of all possible values of x satisfying the equation $2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$, is

   • $3$

   • $\frac{3}{2}$

   • $\frac{5}{2}$

   • $\frac{1}{2}$

   Show solution

   It is given that  $2^{4x^{2}}-2^{2x^{2}+x+16}+2^{2x+30}=0$ , which can be written as:

   => $\left(2^{2x^2}\right)^2-2^{2x^2}\cdot2^{x+15}\cdot2^1+\left(2^{x+15}\right)^{^2}=0$

   =>  $\left(2^{2x^2}-2^{x+15}\right)^{^2}=0$

   =>  $2^{2x^2}-2^{x+15}=0$ (Since  \left(a-b\right)^2\ =\ 0\ =>\ a-b\ =0 )

   =>  $2x^2\ =\ x+15$

   =>  $2x^2-x-15=0$  

   =>  $2x^2-6x+5x-15=0$

   =>  $2x\left(x-3\right)+5\left(x-3\right)=0$

   =>  $\left(2x+5\right)\left(x-3\right)\ =\ 0$

   Hence, the possible values of x are  $-\frac{5}{2}$ , and  $3$ , respectively.

   Therefore, the sum of the possible values is  $\left(3-\frac{5}{2}\right)=\frac{1}{2}$

   The correct option is D

5. Q5.CAT 2022

   Suppose k is any integer such that the equation $2x^{2}+kx+5=0$ has no real roots and the equation $x^{2}+(k-5)x+1=0$ has two distinct real roots for x. Then, the number of possible values of k is

   • 9

   • 7

   • 8

   • 13

   Show solution

   $2x^{2}+kx+5=0$ has no real roots so D<0

   k^2-40\ &lt;0

   \left(k-\sqrt{40}\right)\left(k+\sqrt{40}\right)&lt;0

   $k\in\left(-\sqrt{40},\sqrt{40}\right)$

   $x^{2}+(k-5)x+1=0$ has two distinct real roots so D>0

   \left(k-5\right)^2-4&gt;0

   k^2-10k+21&gt;0

   \left(k-3\right)\left(k-7\right)&gt;0

   $k\in\left(-\infty\ ,3\right)∪\left(7,\infty\ \right)$

   Therefore possibe value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2

   In 9 total 9 integer values of k are possible.

6. Q6.CAT 2021

   If r is a constant such that $\mid x^2 - 4x - 13 \mid = r$ has exactly three distinct real roots, then the value of r is

   • 17

   • 21

   • 15

   • 18

   Show solution

   

   The quadratic equation of the form  $\mid x^2 - 4x - 13 \mid = r$ has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.

   Hence at x = 2 the quadratic equation has its minimum.

   Considering the quadratic part :  $\left|x^2-4\cdot x-13\right|$ . as per the given condition, this must-have 3 real roots.

   The curve ABCDE represents the function  $\left|x^2-4\cdot x-13\right|$ . Because of the modulus function, the representation of the quadratic equation becomes :

   ABC'DE. 

   There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C'.

   The point C' is a reflection of C about the x-axis. r is the y coordinate of the point C' :

   The point C which is the value of the function at x = 2, =  $2^2-8-13$

   = -17, the reflection about the x-axis is 17.

   Alternatively,

   $\mid x^2 - 4x - 13 \mid = r$ .

   This can represented in two parts :

   $x^2-4x-13\ =\ r\ if\ r\ is\ positive.$

   $x^2-4x-13\ =\ -r\ if\ r\ is\ negative.$

   Considering the first case :  $x^2-4x-13\ =r$

   The quadraticequation becomes :  $x^2-4x-13-r\ =\ 0$

   The discriminant for this function is :  $b^2-4ac\ =\ 16-\ \left(4\cdot\left(-13-r\right)\right)=68+4r$

   SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.

   For the second case :

   $x^2-4x-13+r\ =\ 0$ the function inside the modulus is negaitve

   The discriminant is  $16\ -\ \left(4\cdot\left(r-13\right)\right)\ =\ 68-4r$

   In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots.

   Hence  $\ 68-4r\ =\ 0$

   r = 17, for r = 17 we can have exactly 3 roots.

7. Q7.CAT 2020

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

   • 8

   • 4

   • 2

   • 6

   Show solution

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

   if  $(x^{2}-13x+42)$ =0 or  $(x^{2}-7x+11)$ =1 or  $(x^{2}-7x+11)$ =-1 and  $(x^{2}-13x+42)$ 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,  $(x^{2}-13x+42)$ is even number.

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

   .'. x can take six values. 

8. Q8.CAT 2019

   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 &gt; \frac{1}{16}

   • A &lt; \frac{1}{16}

   • A &lt; \frac{1}{8}

   • A &gt; \frac{1}{8}

   Show solution

   The roots of  $x^2 - 4x - log_{2}{A} = 0$ 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

9. Q9.CAT 2018

   If $U^{2}+(U-2V-1)^{2}$= −$4V(U+V)$ , then what is the value of $U+3V$ ?

   • $0$

   • $\dfrac{1}{2}$

   • $\dfrac{-1}{4}$

   • $\dfrac{1}{4}$

   Show solution

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

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

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

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

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

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

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

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

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

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

10. Q10.CAT 2007

    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?

    • -119

    • -159

    • -110

    • -180

    • -105

    Show 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.

11. Q11.CAT 2002

    Davji Shop sells samosas in boxes of different sizes. The samosas are priced at Rs. 2 per samosa up to 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paise per samosa. What should be the maximum size of the box that would maximise the revenue?

    • 240

    • 300

    • 400

    • None of these

    Show solution

    Let the optimum number of samosas be 200+20n

    So, price of each samosa = (2-0.1*n)

    Total price of all samosas = (2-0.1*n)*(200+20n) = $400 - 20n + 40n - 2n^2$ = $400 + 20n - 2n^2$

    This quadratic equation attains a maximum at n = -20/2*(-2) = 5

    So, the number of samosas to get the maximum revenue = 200 + 20*5 = 300

12. Q12.CAT 1996

    Given the quadratic equation $x^2 - (A - 3)x - (A - 2)$, for what value of $A$ will the sum of the squares of the roots be zero?

    • -2

    • 3

    • 6

    • None of these

    Show solution

    For summation of square of roots to be zero, individual roots should be zero.
    Hence summation should be zero i.e.  A-3=0 ; A = 3
    And product of roots will also be zero i.e. A-2 = 0 ; A =2
    So there is no unique value of A which can satisfy above equation.

CAT Quadratic Equations previous year questions with solutions — AthenaPrep