# The Square Root Method and Completing the Square

## The Square Root Method

Suppose you have an equation like $(x + 5)^2 = 14$. With the tools we have so far, your best approach would be to multiply out the left side, set the equation equal to zero and hope that it factors. We can actually solve this a lot quicker than that but we need a new property.

## The Square Root Property

If b is a real number and $a^2 = b$ then $a=\pm \sqrt{b}$.

The way I remember this is just to think about taking the square root of both sides of the equation:

$$a^2 = b$$ $$\sqrt{a^2} = \pm\sqrt{b}$$ $$a = \pm\sqrt{b}$$

The only tricky thing you have to keep in mind is that the side with the square root has to have the "plus minus" symbol or ± in front of it so you're getting all the possible solutions.

# Example 1

Solve the equation $(x + 5)^2 = 14$.

Using our new property (or just taking the square root of both sides - whichever is easier for you to remember), makes the equation

$$x+5=\pm\sqrt{14}$$ $$x=-5\pm\sqrt{14}$$

# Example 2

Solve the equation $(3 -x)^2 = -4$.

$$3-x=\pm\sqrt{-4}$$ $$3-x=\pm2i$$ $$-x=-3\pm2i$$ $$x=3\pm2i$$

On the one hand, this is a really powerful tool to add to our toolbox. If you multiply out the expression in the first example and set it equal to 0, you get

$$x^2 +10x + 11 = 0$$

That equation doesn't factor so, if factoring was our only option, we would have had to stop and say the equation couldn't be solved. This "square root method" has opened up a whole new class of equations that we can solve that we weren't able to before.

The downside is that the equation has to be in a very special form that looks like something squared equals a number. Very few equations are going to come to us in that special form. Fortunately, there's another method we can use that will let us take any quadratic equation and rewrite it into the form we need to use our new tool.

## Completing the Square

What we need to be able to do is take an equation that looks like $ax^2 + bx + c = 0$ and "transform it" into a version that looks like $(x + c)^2 =d$. To make this happen, I need to add something to the equation that makes the $ax^2 + bx$ part into a perfect square. To do that, I'm going to add $\frac{b^2}{4a}$ to both sides. (If you want to see where that number comes from, take a look at the first video in the Videos section below.)

# Example 1

Solve the equation $x^2 + 6x + 4 = 0$.

The first thing I'm going to do is move the 4 to the right side of the equation. I'm thinking ahead and I know that to use the Square Root Method, that's the side where the number needs to be.

$$x^2 + 6x = -4$$

Now, I'm going to add my "special number" to each side. Since a = 1 and b = 6, the number I'm going to add is $\frac{6^2}{4\cdot 1} = 9$.

$$x^2 + 6x + 9 = -4 + 9$$

Now, not only does the left side factor but it factors to a perfect square.

$$(x + 3)^2 = 5$$

Now we can use the Square Root Method to get the final answer.

$$x + 3 = \pm\sqrt{5}$$ $$x = -3 \pm\sqrt{5}$$

# Example 3

Solve the equation $3x^2 - 4x + 5 = 0$.

First, the 5 needs to move to the right side.

$$3x^2 - 4x = -5$$

Since a = 3 and b = -4, the "special number" for this equation is $\frac{(-4)^2}{4\cdot3} = \frac{4}{3}$.

$$3x^2 - 4x + \frac{4}{3} = -5 + \frac{4}{3}$$

The left side is a perfect square but it's a little hard to see. The factorization will be clearer if we multiply both sides by 3 to get rid of the fraction.

$$3(3x^2 - 4x + \frac{4}{3}) = 3 (-5 + \frac{4}{3})$$ $$9x^2 - 12x + 4 = -15 + 4$$ $$(3x+2)^2 = -11$$ $$3x+2 = \pm\sqrt{-11}$$ $$3x+2 = \pm i \sqrt{11}$$ $$3x = -2 \pm i \sqrt{11}$$ $$x = \frac{-2 \pm i \sqrt{11}}{3}$$

# Example 2

Solve the equation $x^2 + 4x + 6 = 0$.

First, the 2 needs to move to the right side.

$$x^2 + 4x = -6$$

Since a = 1 and b = 4, the "special number" for this equation is $(\frac{4}{2\cdot1})^2 = 4$.

$$x^2 + 4x + 4 = -6 + 4$$

Now, not only does the left side factor but it factors to a perfect square.

$$(x + 2)^2 = -2$$

Now we can use the Square Root Method to get the final answer.

$$x + 2 = \pm\sqrt{-2}$$ $$x + 2 = \pm i\sqrt{2}$$ $$x = -2 \pm i\sqrt{3}$$

# Example 4

Solve the equation $3x^2 + 12x + 1 = 0$.

First, I'll move the 1 to the right side of the equation.

$$3x^2 + 12x = -1$$

Since a = 3 and b = 12, the number I'm going to add is $\frac{12^2}{4\cdot 3} = 12$.

$$3x^2 + 12x + 12 = -1 + 12$$ $$3x^2 + 12x + 12 = 11$$

Now, the left side may not look like a perfect square but it will if we factor out the greatest common factor (3).

$$3(x^2 + 4x + 4) = 11$$ $$x^2 + 4x + 4 = \frac{11}{3}$$ $$(x+2)^2 = \frac{11}{3}$$

Now we can use the Square Root Method to get the final answer.

$$x+2 = \pm \sqrt{\frac{11}{3}}$$ $$x = -2 \pm \sqrt{\frac{11}{3}}$$

We're almost done but we've got a little clean up to do. First, I'll rationalize the denominator on the square root part.

$$x = -2 \pm \frac{\sqrt{11}}{\sqrt{3}}$$ $$x = -2 \pm \frac{\sqrt{11}}{\sqrt{3}}\frac{\sqrt{3}}{\sqrt{3}}$$ $$x = -2 \pm \frac{\sqrt{33}}{3}$$

Now we can find a common denominator and add the two parts into a single fraction to get the final answer.

$$x = -\frac{6}{3} \pm \frac{\sqrt{33}}{3}$$ $$x = \frac{-6 \pm \sqrt{33}}{3}$$

On the one hand, this has added a really powerful new tool to our toolbox. We're now in a position where we can solve literally any quadratic equation regardless of whether or not its factorable. The down side is that the method can be a little clunky, especially when there's a number other than 1 in front of the $x^2$ term. In the next section, we're going to use this method to derive a formula that will let us find the solution to any quadratic equation just by substituting numbers into the formula and doing a little simplification.

# Video Lectures

The Square Root Method is a technique for solving a special class of polynomial equations. This may seem really specialized at first but, as we'll see in later lectures, we can often rewrite other polynomials into the form that this method requires making it more generally useful than it looks.
Completing the square is the method that we'll use to convert a generic quadratic equation into one where we can apply our Square Root Method.
In tish lecture, we'll look at some examples of how the completing the square method can be used to solve specific equations. (lecture slides)