How do you solve #x^2-x-1=0# by completing the square?

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Answer 1 · 2018-05-24T11:45:42

We have

#x^2-x-1=0#

#x^2-2*1/2*x+(1/2)^2-5/4=0#

#(x-1/2)^2-5/4=0#

#(x-1/2)^2=5/4#

Taking square roots in both sides

#|x-1/2|=sqrt5/2#

and

#(x-1/2)=+- sqrt5/2#

Hence the two solutions are

#x_1=1/2+sqrt5/2# and #x_2=1/2-sqrt5/2#