How do you solve ${(2 x + 4)}^{2} = 64$ $(2x + 4)^2 = 64$ ?

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Answer 1 · 2015-08-21T07:40:56

The solutions are
$x = - 6, x = 2$ $color(blue)(x=-6,x=2$

${(2 x + 4)}^{2} = 64$ $(2x+4)^2=64$

Taking the square root of both terms:

$\sqrt{{(2 x + 4)}^{2}} = \sqrt{64}$ $sqrt((2x+4)^2)=sqrt64$

$2 x + 4 = \pm 8$ $2x+4=+-8$

Solution 1:
$2 x + 4 = + 8$ $2x+4=+8$

$2 x + 4 - 4 = + 8 - 4$ $2x+4-4=+8-4$

$2 x = 4$ $2x=4$
$x = 2$ $color(blue)(x=2$

Solution 2:
$2 x + 4 = - 8$ $2x+4=-8$

$2 x + 4 - 4 = - 8 - 4$ $2x+4-4=-8-4$

$2 x = - 12$ $2x=-12$
$x = - 6$ $color(blue)(x=-6$

How do you solve (2x+4)2=64(2x + 4)^2 = 64 ?