Question : If $ p(x+y)^{2}=5$ and $q(x-y)^{2}=3$, then the simplified value of $p^{2}(x+y)^{2}+4pq xy -q^{2}(x-y)^{2}$ is:
Option 1: $-(p+q)$
Option 2: $2(p+q)$
Option 3: $p+q$
Option 4: $-2(p+q)$
Correct Answer: $2(p+q)$
Solution :
$p(x+y)^{2}=5$ and $q(x-y)^{2}=3$
$(x+y)^2 = \frac{5}{p}$ and $(x-y)^2 = \frac{3}{q}$
Since $(x+y)^2- (x-y)^2 = 4xy $
So, $ 4xy = \frac{5}{p} - \frac{3}{q}$
By putting it in the given expression, we have,
$p^{2}(x+y)^{2}+4pq xy -q^{2}(x-y)^{2}$
= $p^2 \frac{5}{p} + 4pqxy - q^2 \frac{3}{q}$
= $5p + 4pqxy -3q$
= $5p+pq(\frac{5}{p} - \frac{3}{q})-3q$
= $5p +5q - 3p - 3q = 2(p+q)$
Hence, the correct answer is $2(p+q)$.
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