# How do you solve x^2+5x+2=0 by completing the square?

Apr 17, 2018

$x = - \frac{5}{2} + \frac{\sqrt{17}}{2}$ and $x = - \frac{5}{2} - \frac{\sqrt{17}}{2}$

#### Explanation:

Half the middle term, add it to $x$ and square it. Keep the constant on the end

${x}^{2} + 5 x \to {\left(x + \frac{5}{2}\right)}^{2} + 2$

Take away the term previously halved, but squared.

${\left(x + \frac{5}{2}\right)}^{2} - \frac{25}{4} + \frac{8}{4}$

Turn the extra constant to have the same denominator to make simplifying easier

Simplify:

${\left(x + \frac{5}{2}\right)}^{2} - \frac{17}{4} = 0$

Solving:

Plus the constant, making it cancel out:

${\left(x + \frac{5}{2}\right)}^{2} \cancel{- \frac{17}{4}} = \frac{17}{4}$

Get rid of the square bracket by square rooting:

$x + \frac{5}{2} = \pm \sqrt{\frac{17}{4}}$ $\leftarrow$ Notice the $\pm$ sign, we add this when square rooting.

Minus the constant:

$x + \cancel{\frac{5}{2}} = - \frac{5}{2} \pm \sqrt{\frac{17}{4}}$

Simplify (If needed):

$\sqrt{4} \to 2$

$x = - \frac{5}{2} \pm \frac{\sqrt{17}}{2}$

$x = - \frac{5}{2} + \frac{\sqrt{17}}{2}$ and $x = - \frac{5}{2} - \frac{\sqrt{17}}{2}$