# How do you solve x^2 + 12x + 20 = 0 by completing the square?

Mar 3, 2018

$x = - 10 , x = - 2$

#### Explanation:

Our Quadratic is in the form $a {x}^{2} + b x + c = 0$. When we complete the square, we take half of the $b$ value, square it, and add it to both sides.

First, let's subtract $20$ from both sides. We get:

${x}^{2} + 12 x = - 20$

Half of $12$ is $6$, and when we square $6$, we get $36$. Let's add this to both sides. We get:

${x}^{2} + 12 x + 36 = - 20 + 36$

Simplifying, we get:

${x}^{2} + 12 x + 36 = 16$

Now, we factor the left side of the equation. We think of two numbers that add up to $12$ and have a product of $36$. $6$ and $6$ are our two numbers.

We can now factor this as:

$\textcolor{b l u e}{\left(x + 6\right) \left(x + 6\right)} = 16$

Notice, what I have in blue is the same as ${x}^{2} + 12 x + 36$. We can rewrite this as:

${\left(x + 6\right)}^{2} = 16$

Taking the square root of both sides, we get:

$x + 6 = - 4$ and $x + 6 = 4$

Subtracting $6$ from both sides of the equations, we get:

$x = - 10 , x = - 2$

Mar 3, 2018

$x = - 2$ and $x = - 10$

#### Explanation:

By completing the square, we always half the coefficient of $x$, putting it in brackets and squaring it, as a quadratic is always in the form $a {x}^{2} + b x + c$. Always remember to add the constant of $\left(+ 20\right)$ on the end:

$\therefore$ ${x}^{2} + 12 x \to {\left(x + 6\right)}^{2}$

Also by completing the square, we also take away the squared number of half of the previous coefficient (the number in the brackets).

${\left(x + 6\right)}^{2} - 36 + 20$

Simplifying terms:

$- 36 + 20 = - 16$

Plugging this back in, removing the $- 36$ and $20$:

${\left(x + 6\right)}^{2} - 16$

This is in the completed square form. You can always check this by expanding out:

$\left(x + 6\right) \left(x + 6\right) \to {x}^{2} + 6 x + 6 x + 36 - 16 \to {x}^{2} + 12 x + 20$

$\therefore$ This complete the square form is correct.

Solving; so far we know:

${\left(x + 6\right)}^{2} - 16 = 0$

So we add $16$ to solve this, as we cannot solve when equal to 0.

${\left(x + 6\right)}^{2} = 16$

As we do not want squares brackets, the opposite of squaring is square rooting, and this cancels out the squared brackets:

${\left(x + 6\right)}^{2} = 16 \to x + 6 = \pm \sqrt{16}$

Always remember the $\pm$ when square rooting...

As we want $x$ on its own, we have to $- 6$ to get rid of it, transferring the $- 6$ to the other side of the equation.

$x + 6 = \pm \sqrt{16} \to x = - 6 \pm \sqrt{16}$

Always remember there are mostly two solutions, as there is a plus and minus on the answer.

$\therefore$ our answers are...

$x = - 6 \pm \sqrt{16}$ and $x = - 6 \pm \sqrt{16}$

But using our squared numbers:

${1}^{2} = 1 \times 1 = 1$
${2}^{2} = 2 \times 2 = 4$
${3}^{2} = 3 \times 3 = 9$
${4}^{2} = 4 \times 4 = 16$ $\sqrt{16}$ can be simplified further as it is a squared number...

$x = - 6 \pm \sqrt{16} \to x = - 6 \pm 4$

$\therefore$ the answers are:

$x = - 6 + 4$ and $x = - 6 - 4$ since most of the time we have two solutions.

These simplify to:

$x = - 2$ and $x = - 10$