# How do you solve using the completing the square method r^2 + 14r = -13?

Apr 15, 2017

${r}^{2} + 14 r = - 13$

$\implies {r}^{2} + 14 r + 13 = 0$

We could have a go at factoring that, but completing the square is the next step down the list and is what you are asking for.

It is the Quadratic Equation without having to remember a formula!

So:

${r}^{2} + 14 r + 13$

$= {r}^{2} + \left(\textcolor{red}{2 \times} 7\right) r + 13$

[Remember that: ${\left(a + b\right)}^{2} = {a}^{2} + \textcolor{red}{2} a b + {b}^{2}$]

$= {\left(r + 7\right)}^{2} - {7}^{2} + 13$

$= {\left(r + 7\right)}^{2} - 36$

Now:

${\left(r + 7\right)}^{2} - 36 \textcolor{red}{= 0}$

$\implies {\left(r + 7\right)}^{2} = 36$

$\implies r + 7 = \sqrt{36}$

$\implies r = - 7 \pm 6 = - 1 , - 13$

We can check that:

$\left(r + 1\right) \left(r + 13\right) = {r}^{2} + 13 r + r + 13 \textcolor{b l u e}{= {r}^{2} + 14 r + 13}$

Apr 15, 2017

See the method outlined below:

#### Explanation:

First, write it in general form:

${r}^{2} + 14 r + 13 = 0$

Add bracket around the terms that contain $r$, with the constant term outside the bracket (and factor out the $a$ coefficient if there is one):

$\left({r}^{2} + 14 r\right) + 13 = 0$

Now, take half of the value of $b$ (the coefficient of the term in $r$) and square it. Add this value and subtract it as well, within the bracket.

$\left({r}^{2} + 14 r + 49 - 49\right) + 13 = 0$

Move the subtracted ${\left(\frac{b}{2}\right)}^{2}$ term outside the bracket. (This can require some care if there was an $a$ value (a coefficient of the ${r}^{2}$ term greater than 1.)

$\left({r}^{2} + 14 r + 49\right) - 49 + 13 = 0$

By design, the part on the brackets is a perfect square:

${\left(r + 7\right)}^{2} - 36 = 0$

It is now in standard form, so you are done.

The vertex is at (-7, -36).

Apr 15, 2017

$r = \left\{- 1 , - 13\right\}$

#### Explanation:

${r}^{2} + 14 r = - 13$

$\text{Let us rearrange the equation.}$

${r}^{2} + 14 r + 13 = 0$

$\text{let us add+36 -36}$

${r}^{2} + 14 r \textcolor{red}{+ 36 - 36} + 13 = 0$

${r}^{2} + 14 r + 36 + 13 \textcolor{red}{- 36} = 0$

$\textcolor{g r e e n}{{r}^{2} + 14 r + 49} - \textcolor{red}{36} = 0$

$\textcolor{g r e e n}{{r}^{2} + 14 r + 49} = {\left(r + 7\right)}^{2}$

${\left(r + 7\right)}^{2} - 36 = 0$

${\left(r + 7\right)}^{2} - {6}^{2} = 0$

$\text{let us remember } {a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

$\left(r + 7 - 6\right) \left(r + 7 + 6\right) = 0$

$\left(r + 1\right) \left(r + 13\right) = 0$

$\text{Either (r+1) or (r+13) is equal to zero.}$

$r + 1 = 0 \Rightarrow r = - 1$

$r + 13 = 0 \Rightarrow r = - 13$