# How do you solve the equation by completing the square: x^2 + 2x - 8 = 0?

Jul 12, 2016

$x = 2$ and $x = - 4$

#### Explanation:

Let's rewrite our equation first, before completing the square:

${x}^{2} + 2 x - 8 = 0$

x^2+2x + ? = 8 + ?

To complete the square, we take the coefficient on the $x$-term, namely $2$, divide it by $2$, and square the result, giving us

${\left(\frac{2}{2}\right)}^{2} = {1}^{2} = 1$

By replacing the ? marks with our number $1$, we get

${x}^{2} + 2 x + 1 = 9$

In this case, we are looking for two numbers whose product gives us $1$ and when added together, gives us $2$.

Since $1 \cdot 1 = 1$ and $1 + 1 = 2$, we can see that the numbers are $1$ and $1$, which means that we can rewrite our equation in the following way:

${\left(x + 1\right)}^{2} = 9$

By taking the square root of both sides we get

(x+1) = ± sqrt(9)

x+1 = ± 3

Subtracting $1$ from both sides gives us

x = ± 3 -1

So our solutions are $x = 2$ and $x = - 4$