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

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Answer 1 · 2015-08-21T07:40:56

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

#(2x+4)^2=64#

Taking the square root of both terms:

#sqrt((2x+4)^2)=sqrt64#

#2x+4=+-8#

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

#2x+4-4=+8-4#

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

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

#2x+4-4=-8-4#

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