# How do you solve c^2 + 8c + 2 = 5c + 15 by completing the square?

Apr 25, 2018

#### Answer:

See the Explanation:

#### Explanation:

${c}^{2} + 8 c + 2 = 5 c + 15$
${c}^{2} + 3 c = 13$
${c}^{2} + 2 \left(\frac{3}{2}\right) c = 13$
${c}^{2} + 2 \left(\frac{3}{2}\right) c + {\left(\frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} = 13$
${\left(c + \frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} = 13$
${\left(c + \frac{3}{2}\right)}^{2} = 13 + \frac{9}{4}$
$c + \frac{3}{2} = \pm \sqrt{13 + \frac{9}{4}}$
$c = - \frac{3}{2} \pm \frac{\sqrt{61}}{2}$

Apr 25, 2018

#### Answer:

$c = - \frac{3}{2} \pm \frac{1}{2} \sqrt{61}$

#### Explanation:

$\text{rearrange equation into "color(blue)"standard form}$

$\text{subtract "5c+15" from both sides}$

$\Rightarrow {c}^{2} + 3 c - 13 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{using the method of "color(blue)"completing the square}$

• " the coefficient of the "c^2" term must be 1 which it is"

• " add/subtract "(1/2"coefficient of the c-term")^2" to"
${c}^{2} + 3 c$

${c}^{2} + 2 \left(\frac{3}{2}\right) c \textcolor{red}{+ \frac{9}{4}} \textcolor{red}{- \frac{9}{4}} - 13 = 0$

$\Rightarrow {\left(c + \frac{3}{2}\right)}^{2} - \frac{61}{4} = 0$

$\Rightarrow {\left(c + \frac{3}{2}\right)}^{2} = \frac{61}{4}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\Rightarrow c + \frac{3}{2} = \pm \sqrt{\frac{61}{4}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow c + \frac{3}{2} = \pm \frac{1}{2} \sqrt{61}$

$\text{subtract "3/2" from both sides}$

$\Rightarrow c = - \frac{3}{2} \pm \frac{1}{2} \sqrt{61} \leftarrow \textcolor{red}{\text{exact solutions}}$