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

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**Quadratic Equations CAT Problems:**

**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 **$2(xy + yz + xz)$

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**Solutions:**

**1)** Answer (D)

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.

**2)** Answer (D)

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

**3)** Answer (B)

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.

**4)** Answer (B)

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.

**5)** Answer (C)

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

as $xy+yz+xz = 0$

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

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