# How do you solve p^2 + 3p - 9 = 0 by completing the square?

May 28, 2017

$p = - \frac{3}{2} - \frac{\sqrt{45}}{2}$ and $p = - \frac{3}{2} + \frac{\sqrt{45}}{2}$

#### Explanation:

Step 1. Add and subtract the perfect square term.

The perfect square term starts with the value next to the variable $p$, in this case $3$. First, you cut it in half:

$3 \implies \frac{3}{2}$

Then you square the result

$\frac{3}{2} \implies {\left(\frac{3}{2}\right)}^{2} = \frac{9}{4}$

Then add and subtract this term in the original expression.

${p}^{2} + 3 p + \frac{9}{4} - \frac{9}{4} - 9 = 0$

Step 2. Factor the perfect square.

The terms in $\textcolor{b l u e}{\text{blue}}$ are the perfect square.

$\textcolor{b l u e}{{p}^{2} + 3 p + \frac{9}{4}} - \frac{9}{4} - 9 = 0$

$\textcolor{b l u e}{{\left(p + \frac{3}{2}\right)}^{2}} - \frac{9}{4} - 9 = 0$

Step 3. Simplify the remaining terms in $\textcolor{red}{\text{red}}$.

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

${\left(p + \frac{3}{2}\right)}^{2} \textcolor{red}{- \frac{45}{4}} = 0$

Step 4. Solve for $p$.

${\left(p + \frac{3}{2}\right)}^{2} - \frac{45}{4} = 0$

Add $45 / 4$ to both sides

${\left(p + \frac{3}{2}\right)}^{2} = \frac{45}{4}$

Take $\pm$ the square root of both sides.

$p + \frac{3}{2} = \pm \sqrt{\frac{45}{4}}$

Subtract $3 / 2$ from both sides.

$p = - \frac{3}{2} \pm \frac{\sqrt{45}}{2}$

$p = - \frac{3}{2} - \frac{\sqrt{45}}{2}$ and $p = - \frac{3}{2} + \frac{\sqrt{45}}{2}$

May 28, 2017

$p = - \frac{3}{2} \pm \frac{3 \sqrt{5}}{2}$

#### Explanation:

${p}^{2} + 3 p - 9 = 0$ => add 9 to both sides:
${p}^{2} + 3 p = 9$ => using;${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$
${p}^{2} + 3 p + {\left(\frac{3}{2}\right)}^{2} = 9 + \frac{9}{4}$
${\left(p + \frac{3}{2}\right)}^{2} = \frac{45}{4}$ => take square root of both sides:
$p + \frac{3}{2} = \pm \frac{\sqrt{45}}{2}$ => simplify:
$p = - \frac{3}{2} \pm \frac{3 \sqrt{5}}{2}$