<meta http-equiv="refresh" content="1; url=/nojavascript/"> Simplification of Radical Expressions ( Read ) | Algebra | CK-12 Foundation
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What if you wanted to perform an operation on two radical expressions, like $\sqrt{32x} - \sqrt{8x}$ , in which the numbers under the radical signs were different? How could you find the difference? After completing this Concept, you'll be able to add, subtract, multiply, and divide radical expressions.

### Guidance

When we add and subtract radical expressions, we can combine radical terms only when they have the same expression under the radical sign. This is a lot like combining like terms in variable expressions.

#### Example A

Simplify the following expressions as much as possible.

a.) $4 \sqrt{2} + 5 \sqrt{2}$

b.) $2 \sqrt{3} - \sqrt{2} + 5 \sqrt{3} + 10\sqrt{2}$

Solution

a.) $4 \sqrt{2} + 5 \sqrt{2} = 9 \sqrt{2}$

b.) $2 \sqrt{3} - \sqrt{2} + 5 \sqrt{3} + 10\sqrt{2} = 7 \sqrt{3} + 9 \sqrt{2}$

It’s important to reduce all radicals to their simplest form in order to make sure that we’re combining all possible like terms in the expression. For example, the expression $\sqrt{8} - 2\sqrt{50}$ looks like it can’t be simplified any more because it has no like terms. However, when we write each radical in its simplest form we get $2\sqrt{2} - 10 \sqrt{2}$ , and we can combine those terms to get $-8 \sqrt{2}$ .

#### Example B

Simplify the following expressions as much as possible.

a) $4 \sqrt[3]{128} - \sqrt[3]{250}$

b) $3 \sqrt{x^3} - 4x \sqrt{9x}$

Solution

a) $\text{Re-write radicals in simplest terms:} && & = 4 \sqrt[3]{2 \cdot 64} - \sqrt[3]{2 \cdot 125} = 16 \sqrt[3]{2} - 5 \sqrt[3]{2}\\\text{Combine like terms:} && & = 11 \sqrt[3]{2}$

b) $\text{Re-write radicals in simplest terms:} && 3 \sqrt{x^2 \cdot x} - 12x \sqrt{x} & = 3x \sqrt{x} - 12x \sqrt{x}\\\text{Combine like terms:} && & =-9x \sqrt{x}$

When we multiply radical expressions, we use the “raising a product to a power” rule: $\sqrt[m]{x \cdot y} = \sqrt[m]{x} \cdot \sqrt[m]{y}$ . In this case we apply this rule in reverse.

#### Example C

Simplify the expression $\sqrt{6} \cdot \sqrt{8}$ .

Solution:

$\sqrt{6} \cdot \sqrt{8} = \sqrt{6 \cdot 8} = \sqrt{48}$

Or, in simplest radical form: $\sqrt{48} = \sqrt{16 \cdot 3} = 4 \sqrt{3}.$

We’ll also make use of the fact that: $\sqrt{a} \cdot \sqrt{a} = \sqrt{a^2} = a$ .

When we multiply expressions that have numbers on both the outside and inside the radical sign, we treat the numbers outside the radical sign and the numbers inside the radical sign separately. For example, $a \sqrt{b} \cdot c \sqrt{d} = ac \sqrt{bd}$ .

#### Example D

Multiply the following expressions.

a) $\sqrt{2}\left(\sqrt{3} + \sqrt{5}\right )$

b) $2 \sqrt{x}\left (3 \sqrt{y} - \sqrt{x}\right )$

c) $\left (2 + \sqrt{5}\right )\left (2 - \sqrt{6}\right )$

d) $\left (2 \sqrt{x} + 1\right )\left (5 - \sqrt{x}\right )$

Solution

In each case we use distribution to eliminate the parentheses.

a) $\text{Distribute} \ \sqrt{2} \ \text{inside the parentheses:} && \sqrt{2}\left (\sqrt{3} + \sqrt{5}\right ) & = \sqrt{2} \cdot \sqrt{3} + \sqrt{2} \cdot \sqrt{5}\\\text{Use the raising a product to a power'' rule:} && & = \sqrt{2 \cdot 3} + \sqrt{2 \cdot 5}\\\text{Simplify:} && & =\sqrt{6} + \sqrt{10}$

b) $\text{Distribute} \ 2 \sqrt{x} \ \text{inside the parentheses:} && & =(2 \cdot 3)\left (\sqrt{x} \cdot \sqrt{y}\right ) - 2 \cdot \left ( \sqrt{x} \cdot \sqrt{x}\right )\\\text{Multiply:} && & =6 \sqrt{xy} - 2 \sqrt{x^2}\\\text{Simplify:} && & =6 \sqrt{xy} - 2x$

c) $\text{Distribute:} && (2 + \sqrt{5})(2 - \sqrt{6}) & = (2 \cdot 2) - \left (2 \cdot \sqrt{6}\right ) + \left( 2 \cdot \sqrt{5} \right ) - \left ( \sqrt{5} \cdot \sqrt{6} \right )\\\text{Simplify:}&& & =4 - 2 \sqrt{6} + 2 \sqrt{5} - \sqrt{30}$

d) $\text{Distribute:} && \left (2 \sqrt{x} - 1\right )\left (5 - \sqrt{x}\right ) &=10 \sqrt{x} - 2x - 5 + \sqrt{x}\\\text{Simplify:} && & =11 \sqrt{x} - 2x - 5$

Rationalize the Denominator

Often when we work with radicals, we end up with a radical expression in the denominator of a fraction. It’s traditional to write our fractions in a form that doesn’t have radicals in the denominator, so we use a process called rationalizing the denominator to eliminate them.

Rationalizing is easiest when there’s just a radical and nothing else in the denominator, as in the fraction $\frac{2}{\sqrt{3}}$ . All we have to do then is multiply the numerator and denominator by a radical expression that makes the expression inside the radical into a perfect square, cube, or whatever power is appropriate. In the example above, we multiply by $\sqrt{3}$ :

$\frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$

Cube roots and higher are a little trickier than square roots.

#### Example E

How would we rationalize $\frac{7}{\sqrt[3]{5}}$ ?

Solution:

We can’t just multiply by $\sqrt[3]{5}$ , because then the denominator would be $\sqrt[3]{5^2}$ . To make the denominator a whole number, we need to multiply the numerator and the denominator by $\sqrt[3]{5^2}$ :

$\frac{7}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}} = \frac{7 \sqrt[3]{25}}{\sqrt[3]{5^3}} = \frac{7 \sqrt[3]{25}}{5}$

Trickier still is when the expression in the denominator contains more than one term.

#### Example F

Consider the expression $\frac{2}{2 + \sqrt{3}}$ . We can’t just multiply by $\sqrt{3}$ , because we’d have to distribute that term and then the denominator would be $2 \sqrt{3} + 3$ .

Instead, we multiply by $2 - \sqrt{3}$ . This is a good choice because the product $\left (2 + \sqrt{3}\right )\left (2 - \sqrt{3}\right )$ is a product of a sum and a difference, which means it’s a difference of squares. The radicals cancel each other out when we multiply out, and the denominator works out to $\left (2 + \sqrt{3} \right )\left (2 - \sqrt{3}\right ) = 2^2 - \left ( \sqrt{3}\right )^2 = 4 - 3 = 1$ .

When we multiply both the numerator and denominator by $2 - \sqrt{3}$ , we get:

$\frac{2}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2\left (2 - \sqrt{3}\right )}{4 - 3} = \frac{4 - 2 \sqrt{3}}{1} = 4 - 2 \sqrt{3}$

Now consider the expression $\frac{\sqrt{x} - 1}{\sqrt{x} - 2 \sqrt{y}}$ .

In order to eliminate the radical expressions in the denominator we must multiply by $\sqrt{x} + 2 \sqrt{y}$ .

We get: $\frac{\sqrt{x} - 1}{\sqrt{x} - 2 \sqrt{y}} \cdot \frac{\sqrt{x} + 2 \sqrt{y}}{\sqrt{x} + 2 \sqrt{y}} = \frac{\left (\sqrt{x} - 1\right )\left (\sqrt{x} + 2 \sqrt{y}\right )} {\left (\sqrt{x} - 2 \sqrt{y} \right ) \left ( \sqrt{x} + 2 \sqrt{y} \right )} = \frac{x - 2 \sqrt{y} - \sqrt{x} + 2 \sqrt{xy}}{x - 4y}$

Watch this video for help with the Examples above.

### Vocabulary

• When we multiply radical expressions , we use the “raising a product to a power” rule:

$\sqrt[m]{x \cdot y} = \sqrt[m]{x} \cdot \sqrt[m]{y}$ .

• When we multiply expressions that have numbers on both the outside and inside the radical sign, we treat the numbers outside the radical sign and the numbers inside the radical sign separately:

$a \sqrt{b} \cdot c \sqrt{d} = ac \sqrt{bd}$

### Guided Practice

Simplify the following expressions as much as possible.

a) $4 \sqrt{3} + 2 \sqrt{12}$

b) $10 \sqrt{24} - \sqrt{28}$

Solutions:

a) $\text{Simplify} \ \sqrt{12} \ \text{to its simplest form:} && & =4 \sqrt{3} + 2 \sqrt{4 \cdot 3} = 4 \sqrt{3} + 4 \sqrt{3}\\\text{Combine like terms:} && & =8 \sqrt{3}$

b) $\text{Simplify} \ \sqrt{24} \ \text{and} \ \sqrt{28} \ \text{to their simplest form:} && & =10 \sqrt{6 \cdot 4} - \sqrt{7 \cdot 4} = 20 \sqrt{6} - 2 \sqrt{7}\\\text{There are no like terms.}$

### Practice

Simplify the following expressions as much as possible.

1. $3\sqrt{8} - 6 \sqrt{32}$
2. $\sqrt{180} + \sqrt{405}$
3. $\sqrt{6} - \sqrt{27} + 2 \sqrt{54} + 3 \sqrt{48}$
4. $\sqrt{8x^3} - 4x \sqrt{98x}$
5. $\sqrt{48a} + \sqrt{27a}$
6. $\sqrt[3]{4x^3} + x \cdot \sqrt[3]{256}$

Multiply the following expressions.

1. $\sqrt{6}\left (\sqrt{10} + \sqrt{8}\right )$
2. $\left (\sqrt{a} - \sqrt{b}\right )\left (\sqrt{a} + \sqrt{b}\right )$
3. $\left (2 \sqrt{x} + 5\right )\left (2 \sqrt{x} + 5\right )$

Rationalize the denominator.

1. $\frac{7}{\sqrt{5}}$
2. $\frac{9}{\sqrt{10}}$
3. $\frac{2x}{\sqrt{5x}}$
4. $\frac{\sqrt{5}}{\sqrt{3y}}$
5. $\frac{12}{2 - \sqrt{5}}$
6. $\frac{6 + \sqrt{3}}{4 - \sqrt{3}}$
7. $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{x}}$
8. $\frac{5y}{2 \sqrt{y} - 5}$