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

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Answer 1 · 2018-04-22T14:08:36

#x=2# or #x=-6#

Get rid of the square by square rooting both sides:
#sqrt((x+2)^2)# = #± sqrt(16)#

The square root cancels the square:

#x+2# = #± sqrt(16)#

#± sqrt(16)=+4 or -4#

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

#x+2 = 4#

#x=2#

and

#x+2=-4#

#x=-6#