# How do you complete the square to solve x^2 -4x -5=0?

Apr 25, 2018

$0 = \left({x}^{2} - 4 x\right) - 5 = \left({x}^{2} - 4 x + 4\right) - 4 - 5 = {\left(x - 2\right)}^{2} - 9 \quad$ so

${\left(x - 2\right)}^{2} = 9 \mathmr{and} x - 2 = \setminus \pm \setminus \sqrt{9} \mathmr{and} x = 2 \setminus \pm 3$ so

$x = - 1 \mathmr{and} x = 5$

Apr 25, 2018

$- 1 = x = 5$

#### Explanation:

We can regroup the equation as:

$\left({x}^{2} - 4 x\right) - 5 = 0$

This equation is in the form:

$a {x}^{2} + b x + c = 0$ where $a = 1$. (That is important.)

We divide $b$ by two.

$b = - 4$

$\implies \frac{b}{2} = - 2$ square the answer.

$\implies {\left(- 2\right)}^{2} = 4$

$\implies \left({x}^{2} - 4 x + 4\right) - 5 = 0$ Since we add four to the original equation, we need to adjust it by subtracting it by four.

$\implies \left({x}^{2} - 4 x + 4\right) - 5 - 4 = 0$

$\implies \left({x}^{2} - 4 x + 4\right) - 9 = 0$

Now, note that ${x}^{2} - 4 x + 4$ is in the form

${a}^{2} + 2 a b + {b}^{2}$ where $a = x$ and $b = - 2$

We can rewrite this into the form ${\left(a + b\right)}^{2}$

$\implies {\left(x - 2\right)}^{2} - 9 = 0$

$\implies {\left(x - 2\right)}^{2} = 9$

$\implies \left(x - 2\right) = \pm \sqrt{9}$ don't forget the $\pm$!

$\implies x = \pm 3 + 2$

$\implies - 3 + 2 = x = 3 + 2$

$\implies - 1 = x = 5$