How do you find the solution to the quadratic equation 2x^2 - 2x = 1?

May 28, 2015

Here's how you solve it by completing the square:

$1 = 2 {x}^{2} - 2 x = 2 \left({x}^{2} - x\right) = 2 \left({x}^{2} - x + \frac{1}{4} - \frac{1}{4}\right)$

$= 2 \left({\left(x - \frac{1}{2}\right)}^{2} - \frac{1}{4}\right)$

Divide both ends by $2$ to get

${\left(x - \frac{1}{2}\right)}^{2} - \frac{1}{4} = \frac{1}{2}$

Add $\frac{1}{4}$ to both sides to get:

${\left(x - \frac{1}{2}\right)}^{2} = \frac{3}{4}$

Take the square root of both sides, allowing for both positive and negative possibilities:

$x - \frac{1}{2} = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{\sqrt{4}} = \pm \frac{\sqrt{3}}{2}$

Add $\frac{1}{2}$ to both sides to get:

$x = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$