How do you solve #(x-3)^2=36#?

2 Answers
Oct 31, 2015

Answer:

#x=9 and x=-3#

Explanation:

#(x-3)^2 = 36 #

#(x-3)^2 = 6^2 #

#sqrt ((x-3)^2) # = 6

# +- (x-3) #= 6

either,
taking +,
#+ (x-3) =6#
#x-3=6#
#x=6+3#
So,
#x=9#

Or,
Taking -,
#-(x-3) =6#
#-x +3 =6#
#-x = 6-3#
#-x= 3#
So,
#x= -3#

Oct 31, 2015

Answer:

I found:
#x_1=9#
#x_2=-3#

Explanation:

You can take the square root of both sides and get:
#(x-3)=+-sqrt(36)#
#x=3+-6#
so you have two possibilities:
#x_1=3+6=9#
#x_2=3-6=-3#
Both work when inserted into your original equation.