# How do you solve x^2= 8x - 15 by completing the square?

Jul 11, 2017

Rearranging first

#### Explanation:

${x}^{2} - 8 x + 15 = 0$

${\left(x - 4\right)}^{2} - {4}^{2} + 15 = 0$

${\left(x - 4\right)}^{2} - 1 = 0$

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

$\left(x - 4\right) = \pm 1$

$x = 4 \pm 1$

This gives two values of x, one where you add the 1 and one where you subtract 1

Jul 12, 2017

$x = 3 \mathmr{and} x = 5$

#### Explanation:

${x}^{2} = 8 x - 15$

${x}^{2} - 8 x = - 15 \text{ } \leftarrow$ leave the constant on the right side

Make a perfect square by adding ${\left(\frac{b}{2}\right)}^{2}$ to both sides
$b = - 8 \text{ } \rightarrow {\left(\frac{- 8}{2}\right)}^{2} = 16$

${x}^{2} - 8 x + 16 = - 15 + 16$

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

$x - 4 = \pm 1$

$x - 4 = 1 \text{ } \rightarrow x = 5$

$x - 4 = - 1 \text{ } \rightarrow x = 3$