How do you solve #(x-5)^2=49#?

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Answer 1 · 2016-05-14T09:33:00

#x=12, -2#

#(x-5)^2=49#

We know that #49 = 7^2 " and "49 = -7^2 #

Find the square-root of both sides:

#x-5=+-7#

When #x-5=7#

Add #5# to both sides:

#x-5color(red)(+5)=7color(red)(+5)#

#x=12#

When #x-5=-7#

Add #5# to both sides:

#x-5color(red)(+5)=-7color(red)(+5)#

#x=-2#