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Explanation:

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 = -ba =(A-3)1=(A-3) 

and αβ = Product of the roots = ca=(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).

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