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How do you solve $2 x^{2} - 8 x = - 8$ $2x^2-8x=-8$ ?

1 Answer

May 17, 2016

$x = 2$ $x=2$

Explanation:

If
$XXX 2 x^{3} - 8 x = - 8$ $color(white)("XXX")2x^3-8x=-8$
then
$XXX x^{2} - 4 x = - 4$ $color(white)("XXX")x^2-4x=-4$ (after dividing by $2$ $2$ )

$XXX x^{2} - 4 x + 4 = 0$ $color(white)("XXX")x^2-4x+4=0$ (after adding $4$ $4$ to both sides)

$XXX {(x - 2)}^{2} = 0$ $color(white)("XXX")(x-2)^2=0$ (factoring)

$XXX (x - 2) = 0$ $color(white)("XXX")(x-2)=0$ (since if $a \cdot b = 0$ $a*b=0$ then $a = 0$ $a=0$ or $b = 0$ $b=0$ )

$XXX x = 2$ $color(white)("XXX")x=2$ (after adding $2$ $2$ to both sides)

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