If $x = 2, y = - 1$ , then the value of $x^{2} + 4 x y + 4 y^{2}$ is

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Solution

Substituting the values,

$x^{2} + 4 x y + 4 y^{2} = (2)^{2} + 4 (2) (- 1) + 4 (- 1)^{2} = 4 - 8 + 4 = 0$

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