How do you solve #x^2-2x=5# by completing the square?

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Answer 1 · 2016-07-11T04:11:59

#x = 1 ± sqrt(6)#

Take the coefficient on the #x#-term, namely #-2#, divide by #2#, and square the result, giving you

#(-2/2)^(2) = (-1)^2 = 1#

Thus, we can now replace every #?# mark in the expression

#x^2-2x + ? = 5 + ?#

with the number #1#, giving us

#x^2-2x+1=6#

We'd like two numbers whose product is #1# and when added together gives us the result of #-2# (the number on the #x#-term).

Since #(-1) * (-1) = 1# and #(-1) + (-1) = -2#, we now have our factors and can rewrite our expression in the following way:

#(x-1)(x-1) = 6#

or

#(x-1)^2 = 6#

Taking the square root of both sides of the equation yields

#sqrt((x-1)^2) = ± sqrt(6)#

#(x-1) = ± sqrt(6)#

Adding #1# to both sides gives us our final result of

#x = 1 ± sqrt(6)#