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

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Answer 1 · 2016-03-08T13:52:28

#x=2, -6#

#(x+2)^2=(+-4)^2#
#sqrt((x+2)^2)=sqrt (+-4^2)#
#x+2=+-4#
#x=2, -6#

Answer 2 · 2016-03-08T14:39:53

#x=2 or x=-6#

#(x+2)^2=16#

#x+2=+-4#

#x=2 or x=-6#

Answer 3 · 2016-03-09T14:21:04

#x=2#

#(x+2)^2=16#

Take the Square root of both sides

#rarrsqrt((x+2)^2)=sqrt16#

#rarrx+2=+-4#

#rarrx=+-4-2#

#rArrx=2, -6#

Check

#(2+2)^2=16#

#(4)^2=16#

#16=16# =)

~~~

#(2-6)^2=16#

#(-4)^2=16#

#16=16#