# How do you solve using the completing the square method 2x^2 = 2x + 15?

Apr 29, 2017

Solution : $x = \frac{1}{2} \left(1 + \sqrt{31}\right) \mathmr{and} x = \frac{1}{2} \left(1 - \sqrt{31}\right)$

#### Explanation:

$2 {x}^{2} = 2 x + 15 \mathmr{and} 2 {x}^{2} - 2 x = 15 \mathmr{and} 2 \left({x}^{2} - x\right) = 15 \mathmr{and} \left({x}^{2} - x\right) = \frac{15}{2}$or

$\left({x}^{2} - x + {\left(\frac{1}{2}\right)}^{2}\right) - \frac{1}{4} = \frac{15}{2} \mathmr{and} {\left(x - \frac{1}{2}\right)}^{2} = \frac{15}{2} + \frac{1}{4} \mathmr{and} {\left(x - \frac{1}{2}\right)}^{2} = \frac{31}{4}$ or
$\left(x - \frac{1}{2}\right) = \pm \sqrt{\frac{31}{4}} \mathmr{and} x = \frac{1}{2} \pm \frac{\sqrt{31}}{2} \mathmr{and} x = \frac{1}{2} \left(1 \pm \sqrt{31}\right)$

Solution : $x = \frac{1}{2} \left(1 + \sqrt{31}\right) \mathmr{and} x = \frac{1}{2} \left(1 - \sqrt{31}\right)$ [Ans]