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Question: The number of value of k for which $(x^2 - (k-2)x - 2k)(x^2 + kx + 2k - 4)$ is a perfect squa is...

The number of value of k for which (x2−(k−2)x−2k)(x2+kx+2k−4)(x^2 - (k-2)x - 2k)(x^2 + kx + 2k - 4) is a perfect squa is

A

1

B

0

C

2

D

none of these

Answer

1

Explanation

Solution

Solution:

Factor each quadratic:

f(x)=x2−(k−2)x−2k=(x−k)(x+2)f(x)=x^2-(k-2)x-2k=(x-k)(x+2) g(x)=x2+kx+2k−4=(x+2)(x−(2−k))g(x)=x^2+kx+2k-4=(x+2)(x-(2-k))

Thus,

f(x)g(x)=(x+2)2(x−k)(x−(2−k))f(x)g(x) = (x+2)^2 (x-k)(x-(2-k))

Let

h(x)=(x−k)(x−(2−k))=x2−2x+k(2−k)h(x)=(x-k)(x-(2-k)) = x^2-2x+k(2-k)

For f(x)g(x)f(x)g(x) to be a perfect square, h(x)h(x) must be a perfect square, i.e.,

h(x)=(x−1)2=x2−2x+1.h(x)=(x-1)^2=x^2-2x+1.

Equate constant terms:

k(2−k)=1⇒2k−k2=1⇒k2−2k+1=0⇒(k−1)2=0.k(2-k)=1 \quad \Rightarrow \quad 2k-k^2=1 \quad \Rightarrow \quad k^2-2k+1=0 \quad \Rightarrow \quad (k-1)^2=0.

Thus, k=1k=1.

Answer: 1 (Option (a))

Explanation:

Factorized both quadratics; observed common factor (x+2)(x+2) and deduced that the remaining quadratic must be a perfect square, leading to k=1k=1.