How do you solve $x^{2} + 16 x = 0$ $x^2 + 16x = 0$ by completing the square?

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How do you solve $x^{2} + 16 x = 0$ $x^2 + 16x = 0$ by completing the square?

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Answer 1 · 2015-05-26T13:29:18

If $x^{2} + 16 x$ $x^2 + 16x$ are the first two terms of an squared expression
${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2 = x^2+2ax+a^2$
then $a = 8$ $a= 8$

$x^{2} + 16 x = 0$ $x^2+16x = 0$
can be written as
$x^{2} + 16 x + 8^{2} = 8^{2}$ $x^2+16x+8^2 = 8^2$

#(x+8)^2 = 64

Taking the square root
$x + 8 = \pm 8$ $x+8 = +-8$

and
$x = 0$ $x= 0$ or $x = 16$ $x=16$

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How do you solve $x^{2} + 16 x = 0$ $x^2 + 16x = 0$ by completing the square?

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