How do you solve #(x-5)^2=4#?

1 Answer
georgef
Jul 16, 2016

We must find which numbers #t# satisfy #t^2=4#

Explanation:

If #t^2=4#, then either #t=2# or #t=-2#.

But then we have from the given equation that either #(x-5)=2# or #(x-5)=-2#. Thus we have either:

#x=5+2#, or #x=5-2#. That is either:

#x=7#, or #x=3#, which are the solutions we are looking for.

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