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

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Answer 1 · 2016-09-08T13:24:50

#x = sqrt2+5 = 6.414#

#x= -sqrt2 +5 = 3.586#

Usually we would multiply out the bracket and make the whole equation equal to 0.

However, this is a special case because there is no x-term, so we can leave it as it is and just find the square root of both sides.

#(x-5)^2 = 2#

#x-5 = +-sqrt2#

#x = sqrt2+5 = 6.414#

#x= -sqrt2 +5 = 3.586#