# How do you solve x² - 2x = 15 by completing the square?

Jun 10, 2018

#### Answer:

$x = 5$ or $x = - 3$

#### Explanation:

Although there is a better method, completing the square method is as such:

rearrange the formula to equate 0:

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

${\left(x - 1\right)}^{2} - 1 - 15 = 0$

Rearrange to get:

${\left(x - 1\right)}^{2} = 16$

Square root on both sides to get:
$\left(x - 1\right) = \pm 4$

Therefore adding 1 to both sides gives us either: $x = 5$ or $x = - 3$

Jun 10, 2018

#### Answer:

$x = 5$ and $x = - 3$

#### Explanation:

${x}^{2} - 2 x - 15 = 0$ is in the format of $a {x}^{2} + b x + c = 0$.

As $c$ has already been moved to the left, let's add ${\left(\frac{b}{2}\right)}^{2}$ to both sides and factor:

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

Your value of $b$ is the coefficient before the $x$ in $b x$ (from $a {x}^{2} + b x + c = 0$):

${x}^{2} - 2 x + {\left(- \frac{2}{2}\right)}^{2} = 15 + {\left(- \frac{2}{2}\right)}^{2}$
${x}^{2} - 2 x + {\left(- 1\right)}^{2} = 15 + {\left(- 1\right)}^{2}$
${x}^{2} - 2 x + 1 = 15 + 1$
$\left(x - 1\right) \left(x - 1\right) = 16$
${\left(x - 1\right)}^{2} = 16$

Now solve for $x$:

$x - 1 = \pm \sqrt{16}$
$x - 1 = \pm 4$
$x = 4 + 1$ and $x = - 4 + 1$

Therefore, $x = 5$ and $x = - 3$

Jun 10, 2018

#### Answer:

$x = - 3 , x = 5$

#### Explanation:

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

$\Rightarrow {\left(x - 1\right)}^{2} - 1 - 15$

$\Rightarrow {\left(x - 1\right)}^{2} - 16 = 0$

$\Rightarrow {\left(x - 1\right)}^{2} = 16$

$\Rightarrow x - 1 = \pm \sqrt{16}$

$\Rightarrow x - 1 = \pm 4$

$\Rightarrow x = 1 \pm 4$

$\Rightarrow x = 1 - 4$ or $x = 1 + 4$

$\Rightarrow x = - 3 , x = 5$