CAT 1996 — QA Question 20

If we write the given equation in the conventional form, i.e. ax² + bx + c, a = 1, b = – (A – 3), i.e. (3 – A) and c = –(A – 2), i.e. (2 – A). 

Let the roots of this equation be α and β. So the sum of the squares of the roots = α² + β² = (α + β)² – 2αβ.

Now (α + β) = Sum of the roots = $\frac{-b}{a}$ $=\frac{(A-3)}{1}=(A-3)$ 

and αβ = Product of the roots = $\frac{c}{a}$$=\frac{(2-A)}{1}=$ (2 - A).

Hence, α² + β² = (A – 3)² – 2(2 – A) = A² – 4A + 5 = 0.

Discriminant of this equation is less than 0, hence there no real value of A possible.

Hence, option (d).