Lesson Objectives
• Demonstrate an understanding of radical expressions
• Learn how to multiply two or more radicals
• Learn how to simplify a radical using the product rule for radicals
• Learn how to simplify a radical using the quotient rule for radicals
• Learn how to simplify a radical with variables
• Learn how to simplify higher-level roots

The product rule for radicals tells us the product of two square roots is the square root of the product. Let's suppose we saw the following problem: $$\sqrt{3}\cdot \sqrt{5}$$ Based on our rule, we could place the 3 and 5 under one square root symbol and find their product: $$\sqrt{3}\cdot \sqrt{5}=\sqrt{3 \cdot 5}=\sqrt{15}$$ Let's look at a few examples.
Example 1: Find each product. $$\sqrt{13}\cdot \sqrt{2}$$ We can write one square root symbol and place the two radicands (13 and 2) underneath. These two numbers will be multiplied together: $$\sqrt{13}\cdot \sqrt{2}=\sqrt{13 \cdot 2}=\sqrt{26}$$ Example 2: Find each product. $$\sqrt{6}\cdot \sqrt{7}$$ We can write one square root symbol and place the two radicands (6 and 7) underneath. These two numbers will be multiplied together: $$\sqrt{6}\cdot \sqrt{7}=\sqrt{6 \cdot 7}=\sqrt{42}$$

### Simplifying Square Roots

A square root is simplified when no perfect square factor remains under the radical sign. Let's suppose we saw the following problem: $$\sqrt{8}$$ If we factor 8, we get 23 or 2 • 2 • 2. We know that 4 is a perfect square (2 • 2). Let's use our product rule for radicals to rewrite our problem: $$\sqrt{8}=\sqrt{4 \cdot 2}$$ We can now break this problem up into: $$\sqrt{4}\cdot \sqrt{2}$$ Since the square root of 4 is 2, we can just replace the square root of 4 with a 2: $$2 \cdot \sqrt{2}$$ We can just place the number outside of the square root symbol to imply multiplication: $$2\sqrt{2}$$ Another way to think about this is to factor the radicand (number under the radical symbol): $$\sqrt{8}=\sqrt{2 \cdot 2 \cdot 2}$$ Since the square root of a number is a number that multiplies by itself to give the number, each time we have a pair (two) of factors, we can remove the two factors from the inside of the square root symbol and place one outside: $$\require{color}\sqrt{8}=\sqrt{\colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot 2}=\colorbox{yellow}{2}\sqrt{2}$$ Let's look at some examples.
Example 3: Simplify each. $$\sqrt{72}$$ We will begin by factoring the radicand (number under the radical symbol):
72 = 2 • 2 • 2 • 3 • 3
Each pair of factors can be removed from the inside of the square root symbol and we can place one outside: $$\sqrt{72}$$ $$\sqrt{\colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot 2 \cdot \colorbox{yellow}{3}\cdot \colorbox{yellow}{3}}$$ $$\colorbox{yellow}{2}\cdot \colorbox{yellow}{3}\cdot \sqrt{2}$$ $$6\sqrt{2}$$ Another way to think about this: $$\sqrt{72}=\sqrt{4}\cdot \sqrt{9}\cdot \sqrt{2}$$ We know the square root of 4 is 2 and the square root of 9 is 3. We can replace each: $$\sqrt{4}\cdot \sqrt{9}\cdot \sqrt{2}=2 \cdot 3 \cdot \sqrt{2}$$ $$6\sqrt{2}$$ Example 4: Simplify each. $$\sqrt{26}\cdot \sqrt{39}$$ We can write this problem as: $$\sqrt{26}\cdot \sqrt{39}=\sqrt{26 \cdot 39}$$ Let's factor each number and look for pairs: $$\sqrt{2 \cdot \colorbox{yellow}{13}\cdot 3 \cdot \colorbox{yellow}{13}}$$ $$\colorbox{yellow}{13}\sqrt{2 \cdot 3}$$ $$13 \sqrt{6}$$ Another way to think about this: $$\sqrt{26}\cdot \sqrt{39}=\sqrt{169}\cdot \sqrt{6}$$ We know the square root of 169 is 13. We can replace this: $$\sqrt{169}\cdot \sqrt{6}=13\sqrt{6}$$

When working with division, we have a similar rule. The quotient rule for radicals tells us the square root of the quotient is the quotient of the square roots. Let's suppose we saw the following problem: $$\sqrt{\frac{100}{81}}$$ We can rewrite this as: $$\frac{\sqrt{100}}{\sqrt{81}}$$ We can then simplify. We know the square root of 100 is 10 and the square root of 81 is 9: $$\frac{\sqrt{100}}{\sqrt{81}}=\frac{10}{9}$$ Let's look at a few examples.
Example 5: Simplify each. $$\sqrt{\frac{90}{5}}$$ Let's rewrite our problem as: $$\sqrt{\frac{90}{5}}=\frac{\sqrt{90}}{\sqrt{5}}$$ Now let's simplify each part: $$\frac{\sqrt{90}}{\sqrt{5}}=\frac{\sqrt{\colorbox{yellow}{3}\cdot \colorbox{yellow}{3}\cdot 2 \cdot 5}}{\sqrt{5}}$$ $$\frac{\sqrt{\colorbox{yellow}{3}\cdot \colorbox{yellow}{3}\cdot 2 \cdot 5}}{\sqrt{5}}=\frac{3 \sqrt{2 \cdot 5}}{\sqrt{5}}$$ Square root of 5 over square root of 5 can be canceled: $$\frac{3 \sqrt{2}\cdot \sqrt{5}}{\sqrt{5}}$$ $$\require{cancel}\frac{3 \sqrt{2}\cdot \cancel{\sqrt{5}}}{\cancel{\sqrt{5}}}=3\sqrt{2}$$

To simplify things, we will assume that all variables represent non-negative real numbers. When we simplify with variables, we use the same approach. Let's look at an example.
Example 6: Simplify each. $$\sqrt{44}\cdot \sqrt{55x^2}$$ $$\sqrt{\colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot \colorbox{yellow}{11}\cdot \colorbox{yellow}{11}\cdot 5 \cdot \colorbox{yellow}{x}\cdot \colorbox{yellow}{x}}$$ $$\colorbox{yellow}{2}\cdot \colorbox{yellow}{11}\cdot \colorbox{yellow}{x}\sqrt{5}$$ $$22x \sqrt{5}$$

### Simplify Higher Level Roots

The product and quotient rules for radicals also works for higher-level roots. As an example, if we want to simplify a cube root, we will look for factors that are perfect cubes. Let's suppose we ran into a problem such as: $$\sqrt{16}$$ We can factor the radicand and look for 3 copies of the same number. When we have a cube root, we look for 3 copies. If we had a fourth root, we would look for four copies, and so on and so forth. Once we find our 3 copies, we can remove these from under the radical symbol and place one outside of the radical. $$\sqrt{16}=\sqrt{\colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot 2}$$ $$\sqrt{\colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot 2}=\colorbox{yellow}{2}\sqrt{2}$$ Another way to think about this: $$\sqrt{16}=\sqrt{8}\cdot \sqrt{2}$$ We know the cube root of 8 is 2, so we can replace this: $$\sqrt{8}\cdot \sqrt{2}=2 \sqrt{2}$$ Let's take a look at an example.
Example 7: Simplify each. $$\sqrt{96x^7}$$ Let's begin by factoring 96:
96 = 2 • 2 • 2 • 2 • 2 • 3
We could rewrite 96 as: 25 • 3
We could also rewrite x7 as: x5 • x2 $$\sqrt{96x^7}=\sqrt{2^5}\cdot \sqrt{x^5}\cdot \sqrt{3x^2}$$ We know the fifth root of 32 is 2 and the fifth root of x5 is x. Let's replace each: $$\sqrt{2^5}\cdot \sqrt{x^5}\cdot \sqrt{3x^2}=2x\sqrt{3x}$$

#### Skills Check:

Example #1

Simplify each. $$-6 \sqrt{100 x^{4}y^{4}}$$

A
$$-40y\sqrt{7x}$$
B
$$-60x^{2}y^{2}$$
C
$$-14x^{2}y\sqrt{7}$$
D
$$30x^{2}\sqrt{3y}$$
E
$$-15x^{2}y\sqrt{5}$$

Example #2

Simplify each. $$5\sqrt{-108x^{8}y^{2}}$$

A
$$-15x^{2}\sqrt{4x^{2}y^{2}}$$
B
$$16xy \sqrt{2x^{2}y}$$
C
$$-6y^{2}\sqrt{6x^{2}y^{2}}$$
D
$$-30y^{2}\sqrt{6xy}$$
E
$$-32\sqrt{2x^{2}y^{2}}$$

Example #3

Simplify each. $$4\sqrt{320x}$$

A
$$-4x \sqrt{4x}$$
B
$$16x \sqrt{6}$$
C
$$-8\sqrt{7x^{3}}$$
D
$$8\sqrt{5x}$$
E
$$20\sqrt{15x}$$         