How do you solve $x^{2} + x = 1$ $x^2+x=1$ by completing the square?

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Answer 1 · 2015-05-05T13:07:28

$x^{2} + x = 1$ $x^2+x=1$

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

Since ${(x + a)}^{2} = x^{2} + 2 a x + a^{2}$ $(x+a)^2= x^2+2ax+a^2$ ,

we see that we must have $2 a = 1$ $2a=1$ , so $a = \frac{1}{2}$ $a = 1/2$ .

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

$x^{2} + x + \frac{1}{4} = 1 + \frac{1}{4}$ $x^2+x+1/4=1+1/4$

Factoring the left and simplifying the right, we get:

${(x + \frac{1}{2})}^{2} = \frac{5}{4}$ $(x+1/2)^2=5/4$ .

Now solve by the square root method:

$x + \frac{1}{2} = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}$ $x+1/2 = +-sqrt(5/4) = +-sqrt5/2$

So $x = - \frac{1}{2} \pm \frac{\sqrt{5}}{2}$ $x=-1/2+-sqrt5/2$ , which may also be written:

$x = \frac{- 1 \pm \sqrt{5}}{2}$ $x = (-1+-sqrt5)/2$

How do you solve x^2+x=1x2+x=1x^2+x=1 by completing the square?