How do you solve #(x - 3)^2 = 25#?

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Answer 1 · 2016-04-13T18:10:57

#=>x=8 or x=-2#

Take the square root of both sides giving

#sqrt( (x-3)^2) = +-sqrt(25)#

Technically it should be #+-sqrt( (x-3)^2) = +-sqrt(25)#

But the same answer would be obtained by

#sqrt( (x-3)^2) = +-sqrt(25)#

#=> x-3=+-5#

#=> x=3+-5#

#=>x=8 or x=-2#