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Quadratic equations CAT problems consists of important quadratic equations questions for CAT. This CAT questions will be very useful for quantitative aptitude for CAT.

Taking a free CAT mock test and solving CAT past papers will definitely help you to get good understanding of the simple, linear and quadratic equations CAT questions.

Question 1:

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. -2

B. 3

C. 6

D. None of these

Question 2:

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

A. 3

B. 4

C. 5

D. 6

Question 3:

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. -119

B. -159

C. -110

D. -180

Question 4:

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?

A. $\frac{-1}{\sqrt 3}$

B. -1

C. 0

D. 1

Question 5:

If $xy + yz + zx = 0$, then $(x + y + z)^2$ equals

A. $(x + y)^2 + xz$

B. $(x + z)^2 + xy$

C. $x^2 + y^2 + z^2$

D. $2(xy + yz + xz)$

Solutions:

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.

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

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.

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.
$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$
as $xy+yz+xz = 0$
so equation will be resolved to $x^2 + y^2 + z^2$