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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 $x^2 â€“ (A â€“ 3)x â€“ (A â€“ 2)$, 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

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.Â 3

B.Â 4

C.Â 5

D.Â 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 $x^3 â€“ ax^2 + bx â€“ c = 0$ are three consecutive integers, then what is the smallest possible value of b?

A.Â $\frac{-1}{\sqrt 3}$

B.Â -1

C.Â 0

D.Â 1

Question 5:

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 $2(xy + yz + xz)$

Solutions:

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.

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.

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.

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