# How do you solve using the completing the square method x^2 – 8x + 13 = 0?

Jun 28, 2016

$x = 5.732 \mathmr{and} x = 2.268$

#### Explanation:

The method of completing the square is based on the fact that there is a pattern in the answer of squaring a binomial.

${\left(x - 5\right)}^{2} = {x}^{2} - 10 x + 25 \text{ Note } {\left(\frac{10}{2}\right)}^{2} = + 25$

The square of (half of the coefficient of x), gives the last term.

So let's find the missing term ....

x^2 - 14x + ......color(red)(??).... rArr x^2 - 14x + color(red)(7^2)

x^2 + 22x + .....color(blue)(??).. rArr x^2 + 22x + color(blue)(11^2)

Once you have all three terms, you can write them as ${\left(x \pm \ldots\right)}^{2}$

${x}^{2} - 14 x + \textcolor{red}{49} = {\left(x - \textcolor{red}{7}\right)}^{2} \text{ -14÷2 = -7, or } \sqrt{49} = 7$

${x}^{2} + 22 x + \textcolor{b l u e}{121} = {\left(x + \textcolor{b l u e}{11}\right)}^{2}$

${x}^{2} - 8 x + 13 = 0 \text{ move 13 to the right side}$

${x}^{2} - 8 x + 16 = - 13 + 16 \text{ add 16 to both sides}$

${\left(x - 4\right)}^{2} = 3 \text{ Write as a square}$

$x - 4 = \pm \sqrt{3} \text{ square root both sides}$

$x = \sqrt{3} + 4 \mathmr{and} x = - \sqrt{3} + 4 \text{ solve for x twice}$

$x = 5.732 \mathmr{and} x = 2.268$