How do you solve ${(x + 2)}^{2} = 16$ $(x + 2)^2 = 16$ ?

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Answer 1 · 2018-04-22T14:08:36

$x = 2$ $x=2$ or $x = - 6$ $x=-6$

Get rid of the square by square rooting both sides:
$\sqrt{{(x + 2)}^{2}}$ $sqrt((x+2)^2)$ = $\pm \sqrt{16}$ $± sqrt(16)$

The square root cancels the square:

$x + 2$ $x+2$ = $\pm \sqrt{16}$ $± sqrt(16)$

$\pm \sqrt{16} = + 4 or - 4$ $± sqrt(16)=+4 or -4$

So you have to solve for both $+ 4$ $+4$ and $- 4$ $-4$

$x + 2 = 4$ $x+2 = 4$

$x = 2$ $x=2$

and

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

$x = - 6$ $x=-6$

How do you solve (x+2)2=16 (x + 2)^2 = 16?