How do you solve ${(x + 5)}^{2} = 9$ $(x+5)^2=9$ ?

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Answer 1 · 2016-07-28T10:17:34

$x = - 2, x = - 8$ $x=-2, x=-8$

${(x + 5)}^{2} = 9$ $(x+5)^2=9$

Taking square Root,

$x + 5 = \sqrt{9} = \pm 3$ $x+5=sqrt9=+-3$

$x + 5 = + 3 \Rightarrow x = 3 - 5 = - 2$ $x+5=+3 rArr x=3-5=-2$ , or,

$x + 5 = - 3 \Rightarrow x = - 3 - 5 = - 8$ $x+5=-3 rArr x=-3-5=-8$

These roots satisfy the given eqn.

Hence, $x = - 2, x = - 8$ $x=-2, x=-8$ are the solns.

How do you solve (x+5)2=9(x+5)^2=9?