Let a, b be the solutions of x^2+p x+1=0 and c, d be the solution of x^2+q x+1=0. If (a-c)(b-c) and (a+d)(b+d) are the solution of x^2+a x+ =0, then is equal to

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Correct answer: D. q^2-p^2

Since, a+b=-p, a b=1 .....(i) and c+d=-q, c d=1 Now (a-c)(b-c) and (a+d)(b+d) are the roots of x^2+a x+ =0 aligned & (a-c)(b-c)(a+d)(b+d)= \ & (a b-a c-b c+c^2 ) (a b+a d+b d+d^2 )= \ & \ 1-c(a+b)+c^2 \ \ 1+d(a+b)+d^2 \ = \ & (1+p c+c^2 ) (1-p d+d^2 )= \ & 1-p d+d^2+p c-p^2 c d+p c d^2+c^2-p c^2 d \ & +c^2 d^2= \ & 1-p d+d^2+p c-p^2+p d+c^2-p c+1= \ & [ c d=1] \ & 2+d^2+c^2-p^2= \ & aligned aligned 2 c d+c^2+d^2-p^2 & = \ (c+d)^2-p^2 & = \ q^2-p^2 & = [ (c+d)=-q] aligned

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