$(a x + b y)^{2} = (a x)^{2} + 16 x y + 4 (y)^{2}$ , then $(a + b)^{2}$ is equal to

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Solution

The correct option is D
36

$(a x + b y)^{2} = (a x)^{2} + 2 a b x y + (b y)^{2}$
Given that, $(a x)^{2} + 2 a b x y + (b y)^{2} = (a x)^{2} + 16 x y + 4 (y)^{2}$
$b^{2} = 4, b = \pm 2$
Case1: b =2
2ab = 16
4a = 16
a = 4
$(a + b)^{2} = (4 + 2)^{2} = 6^{2} = 36$
Case2: b= -2
2a(-2) =16
a =-4
So, $(a + b)^{2} = (- 4 - 2)^{2} = 6^{2} = 36$

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