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

## How to Simplify Radicals

### Product Rule for Radicals

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: $$\sqrt{169}\cdot \sqrt{6}=13\sqrt{6}$$

### Quotient Rule for Radicals

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: $$\frac{3 \sqrt{2}\cdot \sqrt{5}}{\sqrt{5}}$$ $$\require{cancel}\frac{3 \sqrt{2}\cdot \cancel{\sqrt{5}}}{\cancel{\sqrt{5}}}=3\sqrt{2}$$

### Simplifying Radicals with Variables

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: $$\sqrt[3]{16}$$ 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. $$\sqrt[3]{16}=\sqrt[3]{\colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot 2}$$ $$\sqrt[3]{\colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot \colorbox{yellow}{2}\cdot 2}=\colorbox{yellow}{2}\sqrt[3]{2}$$ Another way to think about this: $$\sqrt[3]{16}=\sqrt[3]{8}\cdot \sqrt[3]{2}$$ We know the cube root of 8 is 2, so we can replace this: $$\sqrt[3]{8}\cdot \sqrt[3]{2}=2 \sqrt[3]{2}$$ Let's take a look at an example.
Example 7: Simplify each. $$\sqrt[5]{96x^7}$$ 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 $$\sqrt[5]{96x^7}=\sqrt[5]{2^5}\cdot \sqrt[5]{x^5}\cdot \sqrt[5]{3x^2}$$ We know the fifth root of 32 is 2 and the fifth root of x5 is x. Let's replace each: $$\sqrt[5]{2^5}\cdot \sqrt[5]{x^5}\cdot \sqrt[5]{3x^2}=2x\sqrt[5]{3x}$$

#### Skills Check:

Example #1

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

Please choose the best answer.

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. $$5\sqrt[3]{-108x^{8}y^{2}}$$

Please choose the best answer.

A
$$-15x^{2}\sqrt[3]{4x^{2}y^{2}}$$
B
$$16xy \sqrt[3]{2x^{2}y}$$
C
$$-6y^{2}\sqrt[3]{6x^{2}y^{2}}$$
D
$$-30y^{2}\sqrt[3]{6xy}$$
E
$$-32\sqrt[3]{2x^{2}y^{2}}$$

Example #3

Simplify each. $$4\sqrt[6]{320x}$$

Please choose the best answer.

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

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