# [Important] CAT Quadratic Equations Questions PDF

**Quadratic Equations **is one of the important topics in the **CAT ****Quants (Algebra) **section. These questions are usually not very tough, and hence you should not miss out on the questions from Quadratic Equations. Firstly, understand the concept of **Quadratic Equations, including all the Important Formulas**. Solve & practice more questions from CAT Quadratic Equations so that you can solve these questions in the exam. You can check out these CAT Quadratic Equations-based questions from the **CAT Previous year’s papers**. In this post, we will look into some important Quadratic Equations Questions for CAT. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download these Important **CAT Quadratic Equation Questions** (with detailed answers) **PDFs** along with the video solutions below, which are completely Free.

Download Quadratic Equation Questions for CAT

**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 **For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive?

$x^2 – y^2 = 0$

$(x-k)^2 + y^2 = 1$

a)Â 2

b)Â 0

c)Â $\sqrt2$

d)Â $-2\sqrt2$

**4)Â AnswerÂ (C)**

**Solution:**

From 1st equation we know that $(x)^2 = y^2 $

Substituting this in 2nd equation. we get , $2*x^2 – 2*x*k + k^2-1 =0 $ and for unique solution $b^2-4ac=0$ must satisfy.

This is possible only when k = $\sqrt{2}$

**Question 5: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)

**5)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 6:Â **If the equation $x^3 – ax^2 + bx – a = 0$ has three real roots, then it must be the case that,

a)Â b=1

b)Â b $\neq$ 1

c)Â a=1

d)Â a $\neq$ 1

**6)Â AnswerÂ (B)**

**Solution:**

It can be clearly seen that if b=1 then $x^2(x – a) + (x – a) = 0$ an the equation gives only 1 realÂ valueÂ of x

**Question 7:Â **The number of real roots of the equation $A^2/x + B^2/(x-1) = 1$ , where A and B are real numbers not equal to zero simultaneously, is

a)Â 3

b)Â 1

c)Â 2

d)Â Cannot be determined

**7)Â AnswerÂ (D)**

**Solution:**

The given equation can be written as : $A^2 * (x-1) + B^2 * x = x^2 – x$

=> $x^2 + x(-1 – A^2 – B^2) + A^2 = 0$

Discriminant of the equation = $ (-1 – A^2 – B^2) ^2 – 4A^2$

=$ A^4 + B^4 + 1 – 2A^2 + 2B^2 + 2A^2B^2$

= $ A^4 + B^4 + 1 – 2A^2 – 2B^2 + 2A^2B^2 + 4B^2$

= $ (A^2 + B^2 – 1)^2 + 4B^2$

>= 0, 0 when B =0 and A =1

Hence, the number of roots can be 1 or 2.

Option d) is the correct answer.

**Question 8:Â **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?

a)Â 240

b)Â 300

c)Â 400

d)Â None of these

**8)Â AnswerÂ (B)**

**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

**Question 9:Â **The value of $\frac{(1-d^3)}{(1-d)}$ is

a)Â > 1 if d > -1

b)Â > 3 if d > 1

c)Â > 2 if 0 < d < 0.5

d)Â < 2 if d < -2

**9)Â AnswerÂ (B)**

**Solution:**

$\frac{(1-d^3)}{(1-d)}$ = $1+d^2+d$ (where $d \neq 1$)

Let’s say $f(d)=1+d^2+d$

Now $f(d)$ will always be greater than 0 and have its minimum value at d = -0.5. The value is $\frac{3}{4}$.

For $d<-1$ ; $f(d) >1$

$-1<d<0$ ; $\frac{3}{4} <f(d)<1$

$0<d<1$ ; $1<f(d)<3$

$d>1$ ; $f(d)>3$

So, for d > 1, f(d) > 3. Option b) is the correct answer.

**Question 10:Â **The roots of the equation $ax^{2} + 3x + 6 = 0$ will be reciprocal to each other if the value of a is

a)Â 3

b)Â 4

c)Â 5

d)Â 6

**10)Â AnswerÂ (D)**

**Solution:**

If roots of given equation are reciprocal to each other than product of roots should be equal to 1.

i.e. $\frac{6}{a} = 1$

hence a=6

**Question 11:Â **If $xy + yz + zx = 0$, then $(x + y + z)^2$ equals

a)Â $(x + y)^2 + xz$

b)Â $(x + z)^2 + xy$

c)Â $x^2 + y^2 + z^2$

d)Â $2(xy + yz + xz)$

**11)Â AnswerÂ (C)**

**Solution:**

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$

as $xy+yz+xz = 0$

so equation will be resolved to $x^2 + y^2 + z^2$

**Question 12:Â **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?

a)Â -2

b)Â 3

c)Â 6

d)Â None of these

**12)Â AnswerÂ (D)**

**Solution:**

For summation of square of roots to be zero, individual roots should be zero.

Hence summation should be zero i.e. Â 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.

**Question 13: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

**13)Â AnswerÂ (A)**

**Solution:**

$x_1 + x_2 = 2$

and $7x_2 – 4x_1 = 47$

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

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

**Question 14: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 $7+3\sqrt{5}$

**14)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 15: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 $\dfrac{1}{4}$

**15)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 16: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

**16)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 17: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 $A > \frac{1}{8}$

**17)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