# How do you solve  a^2 - 4a - 21 = 0 by completing the square?

May 24, 2016

The solutions are:
color(green)(a = 7 ,  color(green)(a = -3

#### Explanation:

${a}^{2} - 4 a - 21 = 0$

${a}^{2} - 4 a = 21$

To write the Left Hand Side as a Perfect Square, we add 4 to both sides
a^2 -4a + color(green)(4) = 21 + color(green)(4

${a}^{2} - 2 \cdot a \cdot 2 + {2}^{2} = 25$

Using the Identity color(blue)((a-b)^2 = a^2 - 2ab + b^2, we get

${\left(a - 2\right)}^{2} = 25$

$a - 2 = \sqrt{25}$ or $a - 2 = - \sqrt{25}$

$a - 2 = 5$ or $a - 2 = - 5$

$a = 5 + 2$ or $a = - 5 + 2$

color(green)(a = 7 ,  color(green)(a = -3