# How do you solve r^2 - 9r - 38 = -9 by completing the square?

May 19, 2015

Given ${r}^{2} - 9 r - 38 = - 9$

First add 9 to both sides to get:

${r}^{2} - 9 r - 29 = 0$

Then

$0 = {r}^{2} - 9 r - 29$

$= {r}^{2} - 9 r + \frac{81}{4} - \frac{81}{4} - 29$

$= {\left(r - \frac{9}{2}\right)}^{2} - \left(\frac{81}{4} + 29\right)$

$= {\left(r - \frac{9}{2}\right)}^{2} - \left(\frac{81}{4} + \frac{116}{4}\right)$

$= {\left(r - \frac{9}{2}\right)}^{2} - \frac{197}{4}$

Adding $\frac{197}{4}$ to both ends of this equation, we find:

${\left(r - \frac{9}{2}\right)}^{2} = \frac{197}{4}$

So:

$r - \frac{9}{2} = \pm \sqrt{\frac{197}{4}} = \pm \frac{\sqrt{197}}{\sqrt{4}} = \pm \frac{\sqrt{197}}{2}$

Add $\frac{9}{2}$ to both sides to get:

$r = \frac{9}{2} \pm \frac{\sqrt{197}}{2} = \frac{9 \pm \sqrt{197}}{2}$

In general:

$a {r}^{2} + b r + c = a {\left(r + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

which is zero when

$a {\left(r + \frac{b}{2 a}\right)}^{2} = \left({b}^{2} / \left(4 a\right) - c\right) = \frac{{b}^{2} - 4 a c}{4 a}$

and hence

${\left(r + \frac{b}{2 a}\right)}^{2} = \frac{{b}^{2} - 4 a c}{4 {a}^{2}}$

so

$r + \frac{b}{2 a} = \pm \sqrt{\frac{{b}^{2} - 4 a c}{4 {a}^{2}}} = \pm \frac{\sqrt{{b}^{2} - 4 a c}}{\sqrt{4 {a}^{2}}}$

$= \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

Subtracting $\frac{b}{2 a}$ from both sides we get:

$r = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Does that look familiar?