**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 **CAT Previous year papers**. 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.

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**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

**1)Â AnswerÂ (B)**

**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 $\frac{1}{\sqrt 3}$

**2)Â AnswerÂ (B)**

**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

**3)Â AnswerÂ (D)**

**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)

**4)Â AnswerÂ (A)**

**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

**5)Â AnswerÂ (A)**

**Solution:**

$x_1 + x_2 = 2$

and $7x_2 – 4x_1 = 47$

So $x_1 = -3$ and $x_2 = 5$

And $c = x_1 \times x_2 = -15$

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**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 $7+3\sqrt{5}$

**6)Â AnswerÂ (D)**

**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 $\dfrac{1}{4}$

**7)Â AnswerÂ (C)**

**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

**8)Â Answer:Â 36**

**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 $A > \frac{1}{8}$

**9)Â AnswerÂ (A)**

**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

**10)Â AnswerÂ (D)**

**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

**11)Â AnswerÂ (A)**

**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

**12)Â Answer:Â 5**

**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

**13)Â AnswerÂ (D)**

**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

**14)Â Answer:Â 1**

**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

**15)Â AnswerÂ (B)**

**Solution:**

To have real roots the discriminant should be greater than or equal to 0.

So, $m^2-8n\ge0\ \&\ 4n^2-4m\ge0$

=> $m^2\ge8n\ \&\ n^2\ge m$

Since m,n are positive integers the value of m+n will be minimum when m=4 and n=2.

.’. m+n=6.