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

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Answer 1 · 2015-04-04T08:47:21

First, we Transpose the Constant to one side of the equation.
Transposing $-36$ to the other side we get:
$c^2+4c = 36$
Application of $(a+b)^2 = a^2 + 2ab + b^2$
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 $(2)^2 = 4$ . We add this number to BOTH sides of the Equation.
$c^2+4c+4 = 36+4$
$c^2+4c+4 = 40$
The Left Hand side $c^2+4c+4$ is in the form $a^2 + 2ab + b^2$
where $a$ is $c$ , and $b$ is $2$
The equation can be written as
$(c+2)^2 = 40$

So $(c+2)$ can take either $2sqrt10$ or $-2sqrt10$ as a value. That's because squaring both will give us 40.

$c+2 = 2sqrt10$ (or) $c+2 = -2sqrt10$
$c = 2sqrt10-2$ (or) $c = -2sqrt10-2$
$c = 2(sqrt10-1)$ (or) $c = -2(sqrt10+1)$

Solution : $c = 2(sqrt10-1) , -2(sqrt10+1)$
Verify your answer by substituting these values in the Original Equation $c^2- 36 = -4c$ . You will see that the solution is correct.

How do you solve by completing the square: c²-36= -4cc²-36= -4c?