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Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Use the product and quotient properties of radicals.
• Rationalize the denominator.
• Solve real-world problems using square root functions.

Introduction

A radical reverses the operation of raising a number to a power. For example, the square of 4 is $4^2 = 4 \cdot 4 = 16$, and so the square root of 16 is 4. The symbol for a square root is $\sqrt{\;\;}$. This symbol is also called the radical sign.

In addition to square roots, we can also take cube roots, fourth roots, and so on. For example, since 64 is the cube of 4, 4 is the cube root of 64.

$\sqrt[3]{64} = 4 \qquad \text{since} \qquad 4^3 = 4 \cdot 4 \cdot 4 = 64$

We put an index number in the top left corner of the radical sign to show which root of the number we are seeking. Square roots have an index of 2, but we usually don’t bother to write that out.

$\sqrt[2]{36} = \sqrt{36} = 6$

The cube root of a number gives a number which when raised to the power three gives the number under the radical sign. The fourth root of number gives a number which when raised to the power four gives the number under the radical sign:

$\sqrt[4]{81} = 3 \qquad \text{since} \qquad 3^4 = 3 \cdot 3 \cdot 3 \cdot 3 = 81$

And so on for any power we can name.

Even and Odd Roots

Radical expressions that have even indices are called even roots and radical expressions that have odd indices are called odd roots. There is a very important difference between even and odd roots, because they give drastically different results when the number inside the radical sign is negative.

Any real number raised to an even power results in a positive answer. Therefore, when the index of a radical is even, the number inside the radical sign must be non-negative in order to get a real answer.

On the other hand, a positive number raised to an odd power is positive and a negative number raised to an odd power is negative. Thus, a negative number inside the radical sign is not a problem. It just results in a negative answer.

Example 1

a) $\sqrt{121}$

b) $\sqrt[3]{125}$

c) $\sqrt[4]{-625}$

d) $\sqrt[5]{-32}$

Solution

a) $\sqrt{121} = 11$

b) $\sqrt[3]{125} = 5$

c) $\sqrt[3]{-625}$ is not a real number

d) $\sqrt[5]{-32} = -2$

Use the Product and Quotient Properties of Radicals

Radicals can be re-written as rational powers. The radical: $\sqrt[m]{a^n}$ is defined as $a^{\frac{n}{m}}$.

Example 2

Write each expression as an exponent with a rational value for the exponent.

a) $\sqrt{5}$

b) $\sqrt[4]{a}$

c) $\sqrt[3]{4xy}$

d) $\sqrt[6]{x^5}$

Solution

a) $\sqrt{5} = 5^{\frac{1}{2}}$

b) $\sqrt[4]{a} = a^{\frac{1}{4}}$

c) $\sqrt[3]{4xy} = (4xy)^{\frac{1}{3}}$

d) $\sqrt[6]{x^5} = x^{\frac{5}{6}}$

As a result of this property, for any non-negative number $a$ we know that $\sqrt[n]{a^n} = a^{\frac{n}{n}} = a$.

Since roots of numbers can be treated as powers, we can use exponent rules to simplify and evaluate radical expressions. Let’s review the product and quotient rule of exponents.

$\text{Raising a product to a power:} && (x \cdot y)^n & = x^n \cdot y^n\\\text{Raising a quotient to a power:} && \left(\frac{x}{y} \right)^n & = \frac{x^n}{y^n}$

In radical notation, these properties are written as

$\text{Raising a product to a power:} && \sqrt[m]{x \cdot y} & = \sqrt[m]{x} \ \cdot \sqrt[m]{y}\\\text{Raising a quotient to a power:} && \sqrt[m]{\frac{x}{y}} & = \frac{\sqrt[m]{x}}{\sqrt[m]{y}}$

A very important application of these rules is reducing a radical expression to its simplest form. This means that we apply the root on all the factors of the number that are perfect roots and leave all factors that are not perfect roots inside the radical sign.

For example, in the expression $\sqrt{16}$, the number 16 is a perfect square because $16 = 4^2$. This means that we can simplify it as follows:

$\sqrt{16} = \sqrt{4^2} = 4$ Thus, the square root disappears completely.

On the other hand, in the expression $\sqrt{32}$, the number 32 is not a perfect square, so we can’t just remove the square root. However, we notice that $32 = 16 \cdot 2$, so we can write 32 as the product of a perfect square and another number. Thus,

$\sqrt{32} = \sqrt{16 \cdot 2}$ If we apply the “raising a product to a power” rule we get:

$\sqrt{32} = \sqrt{16 \cdot 2} = \sqrt{16} \ \cdot \sqrt{2}$ Since $\sqrt{16} = 4$, we get: $\sqrt{32} = 4 \cdot \sqrt{2} = \underline{\underline{4 \sqrt{2}}}$

Example 3

Write the following expressions in the simplest radical form.

a) $\sqrt{8}$

b) $\sqrt{50}$

c) $\sqrt{\frac{125}{72}}$

Solution

The strategy is to write the number under the square root as the product of a perfect square and another number. The goal is to find the highest perfect square possible; if we don’t find it right away, we just repeat the procedure until we can’t simplify any longer.

a) $\text{We can write} \ 8 = 4 \cdot 2, \ \text{so} && & \sqrt{8} = \sqrt{4 \cdot 2}.\\\text{With the â€œRaising a product to a powerâ€ rule, that becomes} &&& \sqrt{4} \ \cdot \sqrt{2}.\\\text{Evaluate} \ \sqrt{4} \ \text{and weâ€™re left with} &&& \underline{\underline{2\sqrt{2}}}.$

b) $\text{We can write} \ 50 = 25 \cdot 2, \ \text{so:} && \sqrt{50} & = \sqrt{25 \cdot 2}\\\text{Use â€œRaising a product to a powerâ€ rule:} && &=\sqrt{25} \ \cdot \sqrt{2} = \underline{\underline{5 \sqrt{2}}}$

c) $\text{Use â€œRaising a quotient to a powerâ€ rule to separate the fraction:} && \sqrt{\frac{125}{72}} & = \frac{\sqrt{125}}{\sqrt{72}}\\\text{Re-write each radical as a product of a perfect square and another number:} && & = \frac{ \sqrt{25 \cdot 5}}{\sqrt{36 \cdot 2}} = \frac{5 \sqrt{5}}{6 \sqrt{2}}$

The same method can be applied to reduce radicals of different indices to their simplest form.

Example 4

Write the following expression in the simplest radical form.

a) $\sqrt[3]{40}$

b) $\sqrt[4]{\frac{162}{80}}$

c) $\sqrt[3]{135}$

Solution

In these cases we look for the highest possible perfect cube, fourth power, etc. as indicated by the index of the radical.

a) Here we are looking for the product of the highest perfect cube and another number. We write: $\sqrt[3]{40} = \sqrt[3]{8 \cdot 5} = \sqrt[3]{8} \ \cdot \sqrt[3]{5} = 2 \sqrt[3]{5}$

b) Here we are looking for the product of the highest perfect fourth power and another number.

$\text{Re-write as the quotient of two radicals:} && \sqrt[4]{\frac{162}{80}} & = \frac{\sqrt[4]{162}}{\sqrt[4]{80}}\\\text{Simplify each radical separately:} && & = \frac{\sqrt[4]{81 \cdot 2}}{\sqrt[4]{16 \cdot 5}} = \frac{\sqrt[4]{81} \ \cdot \sqrt[4]{2}} {\sqrt[4]{16} \ \cdot \sqrt[4]{5}} = \frac{3 \sqrt[4]{2}}{2 \sqrt[4]{5}}\\\text{Recombine the fraction under one radical sign:} && & = \frac{3}{2} \sqrt[4]{\frac{2}{5}}$

c) Here we are looking for the product of the highest perfect cube root and another number. Often it’s not very easy to identify the perfect root in the expression under the radical sign. In this case, we can factor the number under the radical sign completely by using a factor tree:

We see that $135 = 3 \cdot 3 \cdot 3 \cdot 5 = 3^3 \cdot 5$. Therefore $\sqrt[3]{135} = \sqrt[3]{3^3 \cdot 5} = \sqrt[3]{3^3} \ \cdot \sqrt[3]{5} = 3 \sqrt[3]{5}$.

(You can find a useful tool for creating factor trees at http://www.softschools.com/math/factors/factor_tree/. Click on “User Number” to type in your own number to factor, or just click “New Number” for a random number if you want more practice factoring.)

Now let’s see some examples involving variables.

Example 5

Write the following expression in the simplest radical form.

a) $\sqrt{12x^3 y^5}$

b) $\sqrt[4]{\frac{1250x^7}{405y^9}}$

Solution

Treat constants and each variable separately and write each expression as the products of a perfect power as indicated by the index of the radical and another number.

a) $\text{Re-write as a product of radicals:} && \sqrt{12x^3y^5} & = \sqrt{12} \ \cdot \sqrt{x^3} \ \cdot \sqrt{y^5}\\\text{Simplify each radical separately:} && \left(\sqrt{4 \cdot 3}\right ) \cdot \left( \sqrt{x^2 \cdot x}\right ) \cdot \left (\sqrt{y^4 \cdot y}\right ) & = \left (2 \sqrt{3}\right ) \cdot \left (x \sqrt{x}\right ) \cdot \left (y^2 \sqrt{y}\right )\\\text{Combine all terms outside and inside the radical sign:} && & =2xy^2 \sqrt{3xy}$

b) $\text{Re-write as a quotient of radicals:} && \sqrt[4]{\frac{1250x^7}{405y^9}} & = \frac{\sqrt[4]{1250x^7}}{\sqrt[4]{405y^9}}\\\text{Simplify each radical separately:} && & = \frac{\sqrt[4]{625 \cdot 2} \ \cdot \sqrt[4]{x^4 \cdot x^3}}{\sqrt[4]{81 \cdot 5} \ \cdot \sqrt[4]{y^4 \cdot y^4 \cdot y}} = \frac{5 \sqrt[4]{2} \cdot x \cdot \sqrt[4]{x^3}}{3 \sqrt[4]{5} \cdot y \cdot y \ \cdot \sqrt[4]{y}} = \frac{5x \sqrt[4]{2x^3}}{3y^2 \sqrt[4]{5y}}\\\text{Recombine fraction under one radical sign:} && &= \frac{5x}{3y^2} \sqrt[4]{\frac{2x^3}{5y}}$

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. For example,

$4 \sqrt{2} + 5 \sqrt{2} & = 9 \sqrt{2}\\& \text{or} \\ 2 \sqrt{3} - \sqrt{2} + 5 \sqrt{3} + 10\sqrt{2} & = 7 \sqrt{3} + 9 \sqrt{2}$

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 6

Simplify the following expressions as much as possible.

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

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

Solution

a) $\text{Simplify} \ \sqrt{12} \ \text{to its simplest form:} && & =4 \sqrt{3} + 2 \sqrt{4 \cdot 3} = 4 \sqrt{3} + 6 \sqrt{3}\\\text{Combine like terms:} && & =10 \sqrt{3}$

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

Example 7

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. For example:

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

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 8

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) $\text{Distribute} \ \sqrt{2} \ \text{inside the parentheses:} && \sqrt{2}\left (\sqrt{3} + \sqrt{5}\right ) & = \sqrt{2} \cdot \sqrt{3} + \sqrt{2} \cdot \sqrt{5}\\\text{Use the â€œraising a product to a powerâ€ rule:} && & = \sqrt{2 \cdot 3} + \sqrt{2 \cdot 5}\\\text{Simplify:} && & =\sqrt{6} + \sqrt{10}$ 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. For example, how would we rationalize $\frac{7}{\sqrt[3]{5}}$? 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. For example, 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: $\frac{\sqrt{x} - 1}{\sqrt{x} - 2 \sqrt{y}} \cdot \frac{\sqrt{x} + 2 \sqrt{y}}{\sqrt{x} + 2 \sqrt{y}} = \frac{\left (\sqrt{x} - 1\right )\left (\sqrt{x} + 2 \sqrt{y}\right )} {\left (\sqrt{x} - 2 \sqrt{y} \right ) \left ( \sqrt{x} + 2 \sqrt{y} \right )} = \frac{x - 2 \sqrt{y} - \sqrt{x} + 2 \sqrt{xy}}{x - 4y}$

Solve Real-World Problems Using Radical Expressions

Radicals often arise in problems involving areas and volumes of geometrical figures.

Example 9

A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square feet. Find the dimensions of the pool and the area of the pool.

Solution

Make a sketch:

Let $x =$ the width of the pool. Then:

Area $=$ length $\times$ width

Combined length of pool and walkway $= 2x + 2$

Combined width of pool and walkway $= x + 2$

$\text{Area} = (2x + 2)(x + 2)$

Since the combined area of pool and walkway is $400 \ ft^2$ we can write the equation

$(2x + 2)(x + 2) = 400$

$\text{Multiply in order to eliminate the parentheses:} && 2x^2 + 4x + 2x + 4 & = 400\\\text{Collect like terms:} && 2x^2 + 6x + 4 &= 400\\\text{Move all terms to one side of the equation:} && 2x^2 + 6x - 396 & = 0\\\text{Divide all terms by 2:} && x^2 + 3x - 198 & = 0\\ \text{Use the quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\&& x & = \frac{-3 \pm \sqrt{3^2 - 4(1)(-198)}}{2(1)}\\&& x & = \frac{-3 \pm \sqrt{801}}{2} = \frac{-3 \pm 28.3}{2}\\&& x & = 12.65 \ feet$

(The other answer is negative, so we can throw it out because only a positive number makes sense for the width of a swimming pool.)

Check by plugging the result in the area formula:

Area $= (2(12.65) + 2)(12.65 + 2) = 27.3 \cdot 14.65 = 400 \ ft^2$.

Example 10

The volume of a soda can is $355 \ cm^3$. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.

Solution

Make a sketch:

Let $x =$ the radius of the cylinder base. Then the height of the cylinder is $4x$.

The volume of a cylinder is given by $V = \pi R^2 \cdot h$; in this case, $R$ is $x$ and $h$ is $4x$, and we know the volume is 355.

Solve the equation:

$355 & = \pi x^2 \cdot(4x)\\355 & = 4 \pi x^3\\x^3 & = \frac{355}{4 \pi}\\x & = \sqrt[3]{\frac{355}{4 \pi}} = 3.046 \ cm$

Check by substituting the result back into the formula:

$V = \pi R^2 \cdot h = \pi (3.046)^2 \cdot (4 \cdot 3.046) = 355 \ cm^3$

So the volume is $355 \ cm^3$. The answer checks out.

Review Questions

1. $\sqrt{169}$
2. $\sqrt[4]{-81}$
3. $\sqrt[3]{-125}$
4. $\sqrt[5]{1024}$

Write each expression as a rational exponent.

1. $\sqrt[3]{14}$
2. $\sqrt[4]{zw}$
3. $\sqrt{a}$
4. $\sqrt[9]{y^3}$

Write the following expressions in simplest radical form.

1. $\sqrt{24}$
2. $\sqrt{300}$
3. $\sqrt[5]{96}$
4. $\sqrt{\frac{240}{567}}$
5. $\sqrt[3]{500}$
6. $\sqrt[6]{64x^8}$
7. $\sqrt[3]{48a^3 b^7}$
8. $\sqrt[3]{\frac{16x^5}{135y^4}}$

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. $\frac{5y}{2 \sqrt{y} - 5}$
9. The volume of a spherical balloon is $950 \ cm^3$. Find the radius of the balloon. (Volume of a sphere $= \frac{4}{3} \pi R^3$).
10. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is $180 \ in^2$, what is the width of the frame?

Feb 23, 2012

Apr 02, 2015