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

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Answer 1 · 2015-05-05T13:07:28

#x^2+x=1#

In order for the left side to be a perfect square, it must be the square of some #x+a#.

Since #(x+a)^2= x^2+2ax+a^2#,

we see that we must have #2a=1#, so #a = 1/2#.

This make #a^2=1/4# which I need on the left to make a perfect square. We'll add 1/4 to bot side to get:

#x^2+x+1/4=1+1/4#

Factoring the left and simplifying the right, we get:

#(x+1/2)^2=5/4#.

Now solve by the square root method:

#x+1/2 = +-sqrt(5/4) = +-sqrt5/2#

So #x=-1/2+-sqrt5/2#, which may also be written:

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