# Completing the square

You already know there’s more than one way to solve quadratic equations. Have you tried completing the square yet?

Let’s try it now!

## What does it mean to complete the square?

Completing the square is a method by which the same value is added to and subtracted from an expression in order to write it as a perfect square.

Remember: A quadratic equation is an equation in which the variable is raised to the second power.

We’ll also need a reminder of the standard form of a quadratic equation, which looks like this:

$$ax^2+bx+c=0$$

### Why is completing the square so useful?

Completing the square is another tool in your tool chest for solving quadratic equations. You’ll find that, even beyond quadratic equations, you can work so much more efficiently once you start recognizing which method to use when. Think of it as a fun challenge — see how simple you can make things!

## How to complete the square

Let’s get into some math and learn how to complete the square! The best way to learn something is to see it in action, so we’ll walk through a few examples together.

### Example 1

Solve this quadratic equation by completing the square:

$$x^2+4x{+}{1}=0$$

Move the constant to the right side of the equation and change its sign:

$$x^2+4x={-}{1}$$

To complete the square while preserving the relation between each side of the equation, the same value needs to be added to both sides (remember the addition and subtraction property of equality!):

$$x^2+{4x}+\mathord{?} =-1+\mathord{?}$$

Write the expression as a product with the factors $$2$$ and $$x$$ so that our expression has the same structure as the formula we want to use:

$$x^2+{2\times x\times2}+\mathord{?} =-1+\mathord{?}$$

Since $$2$$ is part of the middle term, add $$2^2$$ to both sides of the equation:

$${x}^2+{2\times {x}\times{2}}+{2^2} =-1+{2^2}$$

Remember the square of the sum formula $${a}^2+2{a}{b}+{b}^2=({a}+{b})^2$$? We’ll use that to factor the expression on the left-hand side of the equation:

$$({x}+{2})^2 =-1+{2^2}$$

Evaluate the power:

$$({x}+{2})^2 =-1+4$$

Calculate the sum:

$$({x}+{2})^2 =3$$

Take the square root of both sides of the equation, and remember to use both positive and negative roots! We can do this because of the rule stating that, if two expressions are equal, their square roots are also equal:

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

Separate the equation into $$2$$ possible cases (one with the minus root and one with the plus root):

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

$${x}{+}{2} =\sqrt{3}$$

Move the constants to the right-hand side of the equation and change their signs:

$${x} =-\sqrt{3}{-}2$$

$${x} =\sqrt{3}{-}2$$

Our equation has $$2$$ solutions:

$$x_1 =-\sqrt{3}-2, ~x_2=\sqrt{3}-2$$

### Example 2

Solve this quadratic equation by completing the square:

$$-2t^{2} + 3t {+} {2} = 0$$

Move the constant to the right side of the equation and change its sign:

$$-2t^{2} + 3t = {-} {2}$$

We want to isolate the unknown variable on one side, so divide both sides of the equation by $$-2$$:

$$t^{2} -\frac32t = 1$$

To complete the square while preserving the relation between the sides of the equation, we need to add the same value to both sides (there’s the addition and subtraction property of equality again):

$$t^{2} -{\frac32t} + \mathord{?} = 1+ \mathord{?}$$

Write the expression as a product with the factors $$2$$ and $$t$$ so that our expression has the same structure as the formula we want to use:

$$t^2-{2\times t\times\frac34}+{\mathord{?}} =1+{\mathord{?}}$$

Since $$\frac34$$ is part of the middle term, add $$\left(\frac34\right)^2$$ to both sides of the equation:

$${t}^2-{2\times {t}\times{\frac34}}+{\left(\frac34\right)^2}=1+\left(\frac34\right)^2$$

The square of the sum formula is $${a}^2+2{a}{b}+{b}^2=({a}+{b})^2$$, so let’s use it to factor the expression on the left-hand side:

$$\left({t}-{\frac34}\right)^2 =1+{\left(\frac34\right)^2}$$

Evaluate the power:

$$\left(t-\frac34\right)^2 =1+{\frac9{16}}$$

Calculate the sum:

$$\left(t-\frac34\right)^2 =\frac{25}{16}$$

Take the square root of both sides of the equation, remembering to use both positive and negative roots. We can do this because of the rule stating if two expressions are equal, then their square roots are also equal:

$$t-\frac34 =\pm\frac{5}{4}$$

Separate the equation into $$2$$ possible cases (one with the minus root and one with the plus root):

$$t{-}{\frac34} =-\frac{5}{4}$$

$$t{-}{\frac34} =\frac{5}{4}$$

Move the constants to the right-hand side of the equation and change their signs:

$$t =-\frac{5}{4}{+}{\frac34}$$

$$t =\frac{5}{4}{+}{\frac34}$$

$$t =-\frac12$$

$$t =2$$

There we go! The equation has $$2$$ solutions:

$$t_1 =-\frac12, ~t_2=2$$

That wasn’t so bad, right? Let’s review the process so you can try this again any time:

## Study summary

1. Move the constant to the right side of the equation.
2. Divide both sides by the constant.
3. To complete the square, add or subtract the same value to both sides.
4. Factor the expression on the left side of the equation.
5. Calculate the sum or difference on the right side of the equation.
6. Solve the equation.

## Do it yourself!

How did it go with the examples? Are you ready to try some on your own? Here are a few practice problems for you to work through:

Solve a quadratic equation by completing the square:

1. $$x^2-4x+3=8$$
2. $$9t^2+6t+5=13$$
3. $$t^2+3t+\frac{1}{4}=7$$
4. $$a^2+\frac{10}{3}a=\frac{11}{9}$$

Solutions:

1. $$x_1=-1, x_2=5$$
2. $$t_1=-\frac43, ~t_2=\frac23$$
3. $$t_1=-\frac92, ~x_2=\frac32$$
4. $$a_1=-\frac{11}3, ~a_2=\frac13$$

If you’re struggling through the solving process, that’s totally okay! Making mistakes can actually help you learn! If you’re getting too stuck or confused, scan the problem using your Photomath app, and we’ll walk you through to the other side!

Here’s a sneak peek of what you’ll see:

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