# How do you solve by completing the square: c²-36= -4c?

##### 1 Answer
Apr 4, 2015
• First, we Transpose the Constant to one side of the equation.
Transposing $- 36$ to the other side we get:
${c}^{2} + 4 c = 36$

• Application of ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$
We look at the Co-efficient of $c$. It's $4$
We take half of this number (including the sign), giving us $2$
We square this value to get ${\left(2\right)}^{2} = 4$. We add this number to BOTH sides of the Equation.
${c}^{2} + 4 c + 4 = 36 + 4$
${c}^{2} + 4 c + 4 = 40$
The Left Hand side ${c}^{2} + 4 c + 4$ is in the form ${a}^{2} + 2 a b + {b}^{2}$
where $a$ is $c$, and $b$ is $2$

• The equation can be written as
${\left(c + 2\right)}^{2} = 40$

So $\left(c + 2\right)$ can take either $2 \sqrt{10}$ or $- 2 \sqrt{10}$ as a value. That's because squaring both will give us 40.

$c + 2 = 2 \sqrt{10}$ (or) $c + 2 = - 2 \sqrt{10}$
$c = 2 \sqrt{10} - 2$ (or) $c = - 2 \sqrt{10} - 2$
$c = 2 \left(\sqrt{10} - 1\right)$ (or) $c = - 2 \left(\sqrt{10} + 1\right)$

• Solution : $c = 2 \left(\sqrt{10} - 1\right) , - 2 \left(\sqrt{10} + 1\right)$

• Verify your answer by substituting these values in the Original Equation ${c}^{2} - 36 = - 4 c$. You will see that the solution is correct.