How do you solve by completing the square $x^{2} + 4 x - 32 = 0$ $x^2+4x-32=0$ ?

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How do you solve by completing the square $x^{2} + 4 x - 32 = 0$ $x^2+4x-32=0$ ?

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Answer 1 · 2015-04-01T13:09:04

I would try:
$x^{2} + 4 x = 32$ $x^2+4x=32$
$x^{2} + 4 x + 4 - 4 = 32$ $x^2+4x+4-4=32$
$x^{2} + 4 x + 4 = 32 + 4$ $x^2+4x+4=32+4$
${(x + 2)}^{2} = 36$ $(x+2)^2=36$
$x + 2 = \pm \sqrt{36}$ $x+2=+-sqrt(36)$
$x = - 2 \pm 6$ $x=-2+-6$
$x_{1} = - 8$ $x_1=-8$
$x_{2} = 4$ $x_2=4$

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How do you solve by completing the square $x^{2} + 4 x - 32 = 0$ $x^2+4x-32=0$ ?

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How do you solve by completing the square $x^{2} + 4 x - 32 = 0$ $x^2+4x-32=0$ ?