How do you solve #\frac { 1} { 2} ( 2x + 4) ^ { 2} = 12#?

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Answer 1 · 2017-06-25T23:06:31

#x=-2+-sqrt6#

#1/2(2x+4)^2=12#

First, start off by dividing #1/2# on both sides.

#(2x+4)^2=24#

Next, place a radical on both sides to cancel out the exponent.

#sqrt((2x+4)^2)=sqrt24#

#2x+4=sqrt24#

Simplify #sqrt24# and factor #2x+4# so you can cancel out common factors. Don't forget to add the #+-#!

#2(x+2)=+-2sqrt6#

#x+2=+-sqrt6#

The last step is to subtract #2# on both sides.

#x=-2+-sqrt6#