# Difference of Two Squares

The difference of two squares is often used to factorize expressions involving quadratics.
In particular quadratic equations looking like: $x^2-k^2 = 0$

or
$ax^2-k^2 = 0$ Remember, the difference of two squares formula was: $\begin{pmatrix}a+b\end{pmatrix}.\begin{pmatrix}a-b\end{pmatrix} = a^2 - b^2$ This can also be read from right to left as: $a^2-b^2 =\begin{pmatrix}a+b\end{pmatrix}.\begin{pmatrix}a-b\end{pmatrix}$ Now that we've made a note of that let's learn the method.

## Factoring & Difference of Two Squares

Say we're asked to factorize, or write in factored form, the following: $x^2 - 4$ The trick is to see that, we can write this expression as the difference of two squares and write: $x^2 - 4 = x^2-2^2$ Consequently, we can write the quadratic in factored form: $x^2 - 2^2 = \begin{pmatrix}x-2\end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix}$ In fact the difference of two squares can be used to factorize any expression looking like $$ax^2-k^2$$.
Indeed we can write: $ax^2 - k^2 = \begin{pmatrix}\sqrt{a}x-k\end{pmatrix}\begin{pmatrix} \sqrt{a}x + k \end{pmatrix}$

## Tutorial

In the following tutorial we learn how to use the difference of two squares to factorize quadratics.

## Exercise 1

Write each of the following quadratics in factored form:

1. $$x^2-16$$
2. $$4p^2 - 25$$
3. $$x^2-9y^2$$
4. $$100m^2-49n^2$$
5. $$25p^2 - 16q^2$$

1. $$x^2-16 = \begin{pmatrix} x - 4 \end{pmatrix}\begin{pmatrix} x + 4 \end{pmatrix}$$

2. $$4p^2 - 25 = \begin{pmatrix} 2p - 5\end{pmatrix}\begin{pmatrix} 2p + 5 \end{pmatrix}$$

3. $$x^2-9y^2 = \begin{pmatrix} x - 3y\end{pmatrix} \begin{pmatrix} x + 3y \end{pmatrix}$$

4. $$100m^2-49n^2 = \begin{pmatrix} 10m - 7n \end{pmatrix} \begin{pmatrix} 10m + 7n \end{pmatrix}$$

5. $$25p^2 - 16q^2 = \begin{pmatrix} 5p - 4q \end{pmatrix} \begin{pmatrix} 5p + 4a \end{pmatrix}$$

## Solving $$x^2 - k = 0$$

Using the method we can solve any quadratic equation of the type: $x^2 - k = 0$ For instance, we can solve $$x^2 - 16 = 0$$, or eve

### Null Factor Law

The null factor law states:

If a product, $$a\times b = 0$$, then either $$a=0$$, or $$b=0$$, or they both equal $$0$$.
Consequently, if we're faced with a product of the type: $\begin{pmatrix} x - a\end{pmatrix}\begin{pmatrix} x + a \end{pmatrix} = 0$ Then: $x - a = 0$
or
$x + a = 0$ $x-a = 0 \quad$

### Two-Step Method for Solving $$x^2 - k = 0$$

Given a quadratic equation of the type $$x^2 - k = 0$$, we can solve this in two steps:

• Step 1: factor the quadratic using the difference of two squares formula; by the end of this step we should be dealing with an equation looking like $$\begin{pmatrix} x - a\end{pmatrix}\begin{pmatrix} x + a \end{pmatrix} = 0$$.

• Step 2: solve the quadratic equation using the null factor law; we'll find $$x=a$$ or $$x = -a$$.

## Tutorial

In the following tutorial we review the ... .

## Exercise 2

Use factorization to solve each of the following quadratic equations:

1. $$x^2 - 4 = 0$$
2. $$x^2 - 49 = 0$$
3. $$2x^2 - 50 = 0$$
4. $$4x^2 - 9 = 0$$
5. $$16x^2 - 81 = 0$$

## Solution Without Working

1. The solutions to $$x^2 - 4=0$$ are:

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

2. The solutions to $$x^2 - 49 = 0$$ are:

$$x = - 7$$ and $$x = 7$$

3. The solutions to $$2x^2 - 50 = 0$$ are:

$$x = -5$$ and $$x = 5$$

4. The solutions to $$4x^2 - 9 = 0$$ are:

$$x = - \frac{3}{2}$$ and $$x = \frac{3}{2}$$

5. The solutions to $$16x^2 - 81 = 0$$ are:

$$x = - \frac{9}{4}$$ and $$x = - \frac{9}{4}$$

## Solution With Working

1. We factor and solve $$x^2 - 4=0$$ in two steps:

• Step 1: we notice that $$x^2 - 4$$ is the difference of two squares.

Indeed $$x^2 - 4 = x^2 - 2^2$$

So we can write it in factored form: $x^2 - 4 = \begin{pmatrix} x - 2 \end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix}$

• Step 2: we use the null product law to solve the factored equation $$\begin{pmatrix} x - 2 \end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix} = 0$$.

That's:

$$\begin{pmatrix} x - 2 \end{pmatrix}\begin{pmatrix} x + 2 \end{pmatrix} = 0$$ if and only if:

$$x - 2 = 0$$ or $$x+2 = 0$$, which leads to:

$$x = 2$$ or $$x = -2$$.

Finally we can state that this quadratic equation has two solutions:

$$x = 2$$ and $$x = -2$$.

2. We factor and solve $$x^2 - 49=0$$ in two steps:

• Step 1: we notice that $$x^2 - 49$$ is the difference of two squares.

Indeed $$x^2 - 49 = x^2 - 7^2$$.

So we can write it in factored form: $x^2 - 49 = \begin{pmatrix} x - 7 \end{pmatrix}\begin{pmatrix} x + 7 \end{pmatrix}$

• Step 2: we use the null product law to solve the factored equation $$\begin{pmatrix} x - 7 \end{pmatrix}\begin{pmatrix} x + 7 \end{pmatrix} = 0$$.

That's:

$$\begin{pmatrix} x - 7 \end{pmatrix}\begin{pmatrix} x + 7 \end{pmatrix} = 0$$ if and only if:

$$x - 7 = 0$$ or $$x+7 = 0$$, which leads to:

$$x = 7$$ or $$x = -7$$.

Finally we can state that this quadratic equation has two solutions:

$$x = 7$$ and $$x = -7$$.

3. We factor and solve $$2x^2 - 50=0$$ in two steps:

Trick: We can start by dividing both sides of this equation by $$2$$. Dividing $$2x^2 - 50 = 0$$ by $$2$$ it leads to: $x^2 - 25 = 0$ We can now solve it in two steps:
• Step 1: we ss that $$x^2 - 25$$ is the difference of two squares.

Indeed $$x^2 - 25 = x^2 - 5^2$$ so we can write it in factored form: $x^2 - 25 = \begin{pmatrix} x - 5 \end{pmatrix}\begin{pmatrix} x + 5 \end{pmatrix}$

• Step 2: we use the null product law to solve the factored equation $$\begin{pmatrix} x - 5 \end{pmatrix}\begin{pmatrix} x + 5 \end{pmatrix} = 0$$.

That's:

$$\begin{pmatrix} x - 5 \end{pmatrix}\begin{pmatrix} x + 5 \end{pmatrix} = 0$$ if and only if:

$$x - 5 = 0$$ or $$x+5 = 0$$, which leads to:

$$x = 5$$ or $$x = -5$$.

Finally we can state that this quadratic equation has two solutions:

$$x = 5$$ and $$x = -5$$.

4. We factor and solve $$4x^2 - 9=0$$ in two steps:

• Step 1: we notice that $$4x^2 - 9$$ is the difference of two squares.

Indeed $$x^2 - 49 = \begin{pmatrix}2x\end{pmatrix}^2 - 3^2$$.

So we can write it in factored form: $x^2 - 49 = \begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 2x + 3 \end{pmatrix}$

• Step 2: we use the null product law to solve the factored equation $$\begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 3x + 3 \end{pmatrix} = 0$$.

That's:

$$\begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 2x + 3 \end{pmatrix} = 0$$ if and only if:

$$2x - 3 = 0$$ or $$2x+3 = 0$$, which leads to:

$$2x = 3$$ or $$2x = -3$$.

Which, in turn, leads to: $$x = \frac{3}{2}$$ or $$x = -\frac{3}{2}$$.

Finally we can state that this quadratic equation has two solutions:

$$x = \frac{3}{2}$$ and $$x = -\frac{3}{2}$$.

5. We factor and solve $$16x^2 - 81=0$$ in two steps:

• Step 1: we notice that $$16x^2 - 81$$ is the difference of two squares.

Indeed $$16x^2 - 81 = \begin{pmatrix}4x\end{pmatrix}^2 - 9^2$$.

So we can write it in factored form: $16x^2 - 81 = \begin{pmatrix} 4x - 9 \end{pmatrix}\begin{pmatrix} 4x + 9 \end{pmatrix}$

• Step 2: we use the null product law to solve the factored equation $$\begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 3x + 3 \end{pmatrix} = 0$$.

That's:

$$\begin{pmatrix} 2x - 3 \end{pmatrix}\begin{pmatrix} 2x + 3 \end{pmatrix} = 0$$ if and only if:

$$2x - 3 = 0$$ or $$2x+3 = 0$$, which leads to:

$$2x = 3$$ or $$2x = -3$$.

Which, in turn, leads to: $$x = \frac{3}{2}$$ or $$x = -\frac{3}{2}$$.

Finally we can state that this quadratic equation has two solutions:

$$x = \frac{3}{2}$$ and $$x = -\frac{3}{2}$$.

6. We factor and solve $$x^2 -5 = 0$$ in two steps:

• Step 1: Using the fact that $$5 = \begin{pmatrix} \sqrt{5} \end{pmatrix}^2$$, we can write $$x^2 - 5$$ as the difference of two squares:

Indeed $$x^2 - 5 = x^2 - \begin{pmatrix} \sqrt{5} \end{pmatrix}^2$$.

So we can write it in factored form: $x^2 - 5 = \begin{pmatrix} x - \sqrt{5} \end{pmatrix} \begin{pmatrix} x + \sqrt{5} \end{pmatrix}$

• Step 2: we use the null product law to solve the factored equation $$\begin{pmatrix} x - \sqrt{5} \end{pmatrix} \begin{pmatrix} x + \sqrt{5} \end{pmatrix} = 0$$.

That's: $$\begin{pmatrix} x - \sqrt{5} \end{pmatrix} \begin{pmatrix} x + \sqrt{5} \end{pmatrix} = 0$$ if and only if:

$$x - \sqrt{5} = 0$$ or $$x + \sqrt{5} = 0$$.

This leads to: $$x = \sqrt{5}$$ or $$x = - \sqrt{5}$$.

Finally we can state that this quadratic equation has two solutions:

$$x = \sqrt{5}$$ and $$x = - \sqrt{5}$$.