# How do you solve p^ { 2} + 2p= - 4?

Jul 19, 2018

Below

#### Explanation:

${p}^{2} + 2 p = - 4$

${p}^{2} + 2 p + 4 = 0$

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

$p = \frac{- 2 \pm \sqrt{4 - 4 \left(1\right) \left(4\right)}}{2}$

$p = \frac{- 2 \pm \sqrt{- 12}}{2}$

Therefore no solution as $\Delta < 0$

If the answer allows for complex numbers, then

$p = \frac{- 2 \pm i \sqrt{12}}{2}$

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

$p = - 1 \pm i \sqrt{3}$

Therefore, ${p}^{2} + 2 p + 4 = 0$ becomes
$\left(p - \left(- 1 + i \sqrt{3}\right)\right) \left(p - \left(- 1 - i \sqrt{3}\right)\right) = 0$

$\left(p + 1 - i \sqrt{3}\right) \left(p + 1 + i \sqrt{3}\right) = 0$