How do you solve using the completing the square method $x^{2} + 8 x = 0$ $x^2 + 8x = 0$ ?

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Answer 1 · 2018-04-30T07:00:52

$x = 4 or - 8$ $x=4 or -8$

$x^{2} + 8 x = 0$ $x^2+8x=0$
you half the coefficient of $x$ $x$ and the take away this number squared.

$\Rightarrow$ $=>$ ${(x + 4)}^{2} - 16 = 0$ $(x+4)^2-16=0$

so the 8 becomes 4 and $4^{2} = 16$ $4^2=16$

$\Rightarrow {(x + 4)}^{2} = 16$ $=> (x+4)^2=16$

$\Rightarrow x + 4 = \pm 4$ $=> x+4=\pm4$

$\Rightarrow x + 4 = 4$ $=>x+4=4$ or $x + 4 = - 4$ $x+4=-4$

$\Rightarrow x = 0 or x = - 8$ $=> x=0 or x=-8$

How do you solve using the completing the square method x2+8x=0x^2 + 8x = 0?