How do you complete the square to solve $x^{2} + 4 x = 12$ $x^2 + 4x = 12$ ?

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How do you complete the square to solve $x^{2} + 4 x = 12$ $x^2 + 4x = 12$ ?

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Answer 1 · 2015-05-31T14:53:24

$x^{2} + 4 x = 12$ $x^2 + 4x = 12$

To write the Left Hand Side as a Perfect Square, we add 4 to both sides:

$x^{2} + 4 x + 4 = 12 + 4$ $x^2 + 4x +4= 12+4$
$x^{2} + 2 \cdot 2 \cdot x + 2^{2} = 16$ $x^2 + 2*2*x + 2^2 = 16$
Using the Identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $color(blue)((a+b)^2 = a^2 + 2ab + b^2$ , we get
${(x + 2)}^{2} = 16$ $(x+2)^2 = 16$
$x + 2 = + \sqrt{16}$ $x + 2 = +sqrt16$ or $x + 2 = - \sqrt{16}$ $x +2 = -sqrt16$
$x = + 4 - 2 or - 4 - 2$ $x = + 4 -2 or -4 -2$
$x = 2 or x = - 6$ $color(blue)(x = 2 or x=-6$

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How do you complete the square to solve $x^{2} + 4 x = 12$ $x^2 + 4x = 12$ ?

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