# How do you complete the square to solve x^2 + 5x + 6 = 0?

Jun 16, 2015

You can complete the square by first getting the form ${x}^{2} + k x = h$.

${x}^{2} + 5 x = - 6$

Then just add and subtract a certain value that is equal to ${\left(\frac{k}{2}\right)}^{2}$. Just remember that the function is ${x}^{2} + k x$, so $k$ may be negative, but the added ${\left(\frac{k}{2}\right)}^{2}$ will always be positive.

${x}^{2} + 5 x + {\left(\frac{5}{2}\right)}^{2} = {\left(\frac{5}{2}\right)}^{2} - 6$

${\left(x + \frac{5}{2}\right)}^{2} = \frac{25}{4} - \frac{24}{4} = \frac{1}{4}$

${\left(x + \frac{5}{2}\right)}^{2} - \frac{1}{4} = 0$

graph{(x+5/2)^2 - 1/4 [-10, 10, -5, 5]}

If you wanted to solve this:

${\left(x + \frac{5}{2}\right)}^{2} = \frac{1}{4}$

$x + \frac{5}{2} = \pm \sqrt{\frac{1}{4}}$

Thus:
$x + \frac{5}{2} = \pm \sqrt{\frac{1}{4}}$

$x = \pm \left(\sqrt{\frac{1}{4}}\right) - \frac{5}{2}$

$x = \pm \left(\frac{1}{2}\right) - \frac{5}{2}$

$x = \frac{1}{2} - \frac{5}{2} = - 2$

$x = - \frac{1}{2} - \frac{5}{2} = - 3$

${\left(- 2\right)}^{2} + 5 \left(- 2\right) + 6 = 4 - 10 + 6 = 0$
${\left(- 3\right)}^{2} + 5 \left(- 3\right) + 6 = 9 - 15 + 6 = 0$

You could just factor, though...

$\left(x + 2\right) \left(x + 3\right) = {x}^{2} + 2 x + 3 x + 6 = {x}^{2} + 5 x + 6$