Question

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

Answer 1

`x=-10, x=-2`

Explanation:

Our Quadratic is in the form `ax^2+bx+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+12x =-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+12x+36=-20+36`

Simplifying, we get:

`x^2+12x+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:

`color(blue)((x+6)(x+6))=16`

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

`(x+6)^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`

Answer 2

`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 `ax^2+bx+c`. Always remember to add the constant of `(+20)` on the end:

`therefore` `x^2+12x -> (x+6)^2`

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

`(x+6)^2-36+20`

Simplifying terms:

`-36+20=-16`

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

`(x+6)^2-16`

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

`(x+6)(x+6) -> x^2+6x+6x+36-16 -> x^2+12x+20`

`therefore` This complete the square form is correct.

Solving; so far we know:

`(x+6)^2-16=0`

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

`(x+6)^2=16`

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

`(x+6)^2=16 -> x+6=pmsqrt16`

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=pmsqrt16 -> x=-6pmsqrt16`

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

`therefore` our answers are...

`x=-6pmsqrt16` and `x=-6pmsqrt16`

But using our squared numbers:

`1^2=1xx1=1`
`2^2=2xx2=4`
`3^2=3xx3=9`
`4^2=4xx4=16` `sqrt16` can be simplified further as it is a squared number...

`x=-6pmsqrt16 -> x=-6pm4`

`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`