How do you solve #2(x-3)^2=8#?

2 Answers
Jul 11, 2017

Answer:

#x=5# or #x=1#

See below

Explanation:

#2(x−3)^2=8#

Divide both sides by 2: #(x-3)^2=4#

Take the square root of each side:

#(x-3)=+-2#

#x-3=+2" "rarrx=5#
or
#x-3=-2" "rarrx=1#

#x=5# or #x=1#

Jul 11, 2017

Answer:

The answer is x= 5 and 1

Explanation:

First you start by evualting the #2(x-3)^2# part of the equation.
#2(x-3)^2#=8
#2(x^2-6x+9)=8#
#2x^2-12x+18=8#
Then you move over the constant on the right side to the left .
#2x^2-12x+18-8=0#
#2x^2-12x+10=0#
Then divide the whole equation by 2.
#(2x^2-12x+10)/2=0/2#
#x^2-6x+5=0#
Then factor the equation normally.
#(x-5)(x-1)=0#