# How do you solve by completing the square for 5-4x-x^2?

May 30, 2015

I assume you meant to solve $5 - 4 x - {x}^{2} = 0$

This is equivalent to
$\textcolor{w h i t e}{\text{XXXXX}}$${x}^{2} + 4 x = 5$
Completing the square
$\textcolor{w h i t e}{\text{XXXXX}}$${x}^{2} + 4 x + 4 = 5 + 4$
$\textcolor{w h i t e}{\text{XXXXX}}$${\left(x + 2\right)}^{2} = 9$
Take the square root
$\textcolor{w h i t e}{\text{XXXXX}}$$x + 2 = \pm 3$
Solve for x
$\textcolor{w h i t e}{\text{XXXXX}}$$x = - 2 \pm 3$

$x = - 5$ or $x = + 1$