In this video, we are going to look at how to solve equations by completing the square.

For example:
If we are given $x^2+4x-11=0$
we can try to factor it, but it would not work. This is when we would solve by completing the square. First, we want to get rid of the last term, so we will add 11 to both sides, giving us $x^2 + 4x = 11$
Now we can complete the square. $(\frac{b}{2})^2=(\frac{4}{2})^2=4$
Since we want to add 4 to the left side to complete the square, we must also add 4 to the right side to keep the equation the same. At this point we have $x^2 + 4x + 4 = 15$
Now if we factor it, we have $(x+2)(x+2)=15$
Or it can also be written as $(x+2)^2=15$
From here we take the square root of both sides $\sqrt{(x+2)^2}=\sqrt{15}$
So $x+2=\pm\sqrt{15}$
Solve for x by subtracting 2 and the final answer is $x=-2\pm\sqrt{15}$

## Video-Lesson Transcript

Let’s go over solving equations by completing the square.

Just to recap, if we have this: $(x + 3)^2 = 4$

We can solve by getting their square roots. $\sqrt{(x + 3)^2} = \sqrt{4}$ $x + 3 = \pm2$

To solve for $x$, $x + 3 - 3 = - 3 \pm2$ $x = -3 \pm2$ $x = -3 + 2 = -1$
and $x = -3 - 2 = -5$

Our solution set is

{ $-1, -5$}

But sometimes we are not given equations in this form.

But we could translate or rewrite quadratic equations into this form and then use this method.

The way to do that is to complete the square. $x^2 + 4x - 11 = 0$

Let’s try to factor this.

But there are no two numbers that will add up to $4$ and when multiplied the answer is $-11$.

So we cannot factor this one.

That’s why we’re going to solve it using completing the square.

We have to look at $\bigg( \dfrac{b}{2} \bigg)^2$

to determine what number needs to be added here to complete the square.

The first thing we need to do is to get rid of $- 11$. $x^2 + 4x - 11 + 11 = 0 + 11$ $x^2 + 4x = 11$

Now let’s go back to the formula since we have $b$-term $\bigg( \dfrac{b}{2} \bigg)^2$ $\bigg( \dfrac{4}{2} \bigg)^2$ $(2)^2$ $4$

So we’re going to add $4$ on both sides of our equation. $x^2 + 4x = 11$ $x^2 + 4x + 4 = 11 + 4$ $x^2 + 4x + 4 = 15$

Now, we can factor this. $x^2 + 4x + 4 = 15$ $(x + 2) (x + 2) = 15$ $(x + 2)^2 = 15$ $\sqrt{(x + 2)^2} = \sqrt{15}$ $x + 2 = \pm \sqrt{15}$

Now let’s solve for $x$: $x + 2 - 2 = -2 \pm \sqrt{15}$ $x = -2 \pm \sqrt{15}$

So our solution set is

{ $-2 + \sqrt{15}, -2 - \sqrt{15}$}

It’s possible that when we solve the square root we’ll have an answer like this. Where there’s an integer plus or minus a radical term.

If these came out as integers and not this type of terms, that means we could have factored it in the beginning.

Let’s have an example like that. $x^2 + 5x + 6 = 0$

Let’s factor this: $(x + 2) (x + 3) = 0$

Let’s solve it $x + 2 = 0$ $x + 2 - 2 = 0 - 2$ $x = - 2$

and $x + 3 = 0$ $x + 3 - 3 = 0 - 3$ $x = - 3$

Our solution set is

{ $-2, -3$}

This is our previous lesson.

Let’s solve this by completing the square. $x^2 + 5x + 6 = 0$ $x^2 + 5x + 6 - 6 = 0 - 6$ $x^2 + 5x = -6$

Let’s now use the formula $\bigg( \dfrac{b}{2} \bigg)^2$ $\bigg( \dfrac{5}{2} \bigg)^2$ $\dfrac{25}{4}$

Let’s go back to solving and add $\dfrac{25}{4}$ in the equation $x^2 + 5x + \dfrac{25}{4} = -6 + \dfrac{25}{4}$

Let’s solve the terms on the right side. $\dfrac{-6}{1} + \dfrac{25}{4}$ $\dfrac{-6(4)}{1(4)} + \dfrac{25}{4}$ $\dfrac{-24}{4} + \dfrac{25}{4}$ $\dfrac{1}{4}$

Now, let’s factor the equation $x^2 + 5x + \dfrac{25}{4} = \dfrac{1}{4}$ $\bigg( x + \dfrac{5}{2} \bigg) \bigg( x + \dfrac{5}{2} \bigg) = \dfrac{1}{4}$ $\bigg( x + \dfrac{5}{2} \bigg)^2 = \dfrac{1}{4}$

Then let’s find its square root $\sqrt{\bigg( x + \dfrac{5}{2} \bigg)^2} = \sqrt{\dfrac{1}{4}}$ $x + \dfrac{5}{2} = \pm \dfrac{1}{2}$

Then let’s solve for $x$: $x + \dfrac{5}{2} - \dfrac{5}{2} = \pm \dfrac{1}{2} - \dfrac{5}{2}$ $x = - \dfrac{5}{2} \pm \dfrac{1}{2}$

So we have two solutions $-\dfrac{5}{2} + \dfrac{1}{2}, -\dfrac{5}{2} - \dfrac{1}{2}$

or { $-2, -3$}

Our answer is the same when we did factoring earlier.

In this case, we could agree that it’s easier to factor than to complete the square.

But of course, the rule for completing the square is true.