# How do you solve k^2 + 6k - 24 = 0 by completing the square?

Apr 6, 2016

The solutions are:
color(green)(k = sqrt 33 - 3 or ,  color(green)(k = -sqrt 33 -3

#### Explanation:

${k}^{2} + 6 k - 24 = 0$

${k}^{2} + 6 k = 24$

To write the Left Hand Side as a Perfect Square, we add 9 to both sides
${k}^{2} + 6 k + 9 = 24 + 9$

${k}^{2} + 2 \cdot k \cdot 3 + {3}^{2} = 33$

Using the Identity color(blue)((a+b)^2 = a^2 + 2ab + b^2, we get
${\left(k + 3\right)}^{2} = 33$

$k + 3 = \sqrt{33}$ or $k + 3 = - \sqrt{33}$

color(green)(k = sqrt 33 - 3 or  color(green)(x = -sqrt 33 -3