# How do you solve using the completing the square method y^2+4y-45=0?

Nov 23, 2016

$y = - 9 \mathmr{and} y = 5$

#### Explanation:

To complete the square we use $\frac{1}{2}$ of the $y$ coefficient as follows:

${y}^{2} + 4 y - 45 = 0$
$\therefore {\left({y}^{2} + \frac{1}{2} \left(4\right)\right)}^{2} - {\left(\frac{1}{2} \left(4\right)\right)}^{2} - 45 = 0$
$\therefore {\left({y}^{2} + 2\right)}^{2} - {\left(2\right)}^{2} - 45 = 0$
$\therefore {\left({y}^{2} + 2\right)}^{2} - 4 - 45 = 0$
$\therefore {\left({y}^{2} + 2\right)}^{2} - 49 = 0$
$\therefore {\left({y}^{2} + 2\right)}^{2} = 49$

We can now take the square root remembering that when we do so we get two solutions as ${a}^{2} = b \iff a = \pm \sqrt{b}$

$\therefore {y}^{2} + 2 = \pm 7$
$\therefore {y}^{2} = - 2 \pm 7$

Yielding the two solutions:

$y = - 9 \mathmr{and} y = 5$