A radical, or root, is the mathematical opposite of an exponent, in the same sense that addition is the opposite of subtraction. The smallest radical is the square root, represented with the symbol √. The next radical is the cube root, represented by the symbol ³√. The small number in front of the radical is its index number. The index number can be any whole number and it also represents the exponent that could be used to cancel out that radical. For example, raising to the power of 3 would cancel out a cube root.

The result of a radical operation is positive if the number under the radical is positive. The result is negative if the number under the radical is negative and the index number is odd. A negative number under the radical with an even index number produces an irrational number. Remember that though it isn't shown, the index number of a square root is 2.

Product and Quotient Rules

To multiply or divide two radicals, the radicals must have the same index number. The product rule dictates that the multiplication of two radicals simply multiplies the values within and places the answer within the same type of radical, simplifying if possible. For example,

\sqrt[3]{2}× \sqrt[3]{4} = \sqrt[3]{8}

which can be simplified to 2. This rule can also work in reverse, splitting a larger radical into two smaller radical multiples.

The quotient rule states that one radical divided by another is the same as dividing the numbers and placing them under the same radical symbol. For example,

\frac{\sqrt{4}}{\sqrt{8}} = \sqrt{\frac{4}{8}} = \sqrt{\frac{1}{2}}

Just like the product rule, you can also reverse the quotient rule to split a fraction under a radical into two individual radicals.

Simplifying Square Roots and Other Radicals

Some radicals solve easily as the number inside solves to a whole number, such as √16 = 4. But most won't simplify as cleanly. The product rule can be used in reverse to simplify trickier radicals. For example, √27 also equals √9 × √3. Since √9 = 3, this problem can be simplified to 3√3. This can be done even when a variable is under the radical, though the variable has to remain under the radical.

Rational fractions can be solved similarly using the quotient rule. For example,

\sqrt{\frac{5}{49}} = \frac{\sqrt{5}}{\sqrt{49}}

Since √49 = 7, the fraction can be simplified to √5 ÷ 7.

Exponents, Radicals and Simplifying Square Roots

Radicals can be eliminated from equations using the exponent version of the index number. For example, in the equation √​x​ = 4, the radical is canceled out by raising both sides to the second power:

(\sqrt{x})^2 = (4)^2\text{ or } x = 16

The inverse exponent of the index number is equivalent to the radical itself. For example, √9 is the same as 91/2. Writing the radical in this manner may come in handy when working with an equation that has a large number of exponents.

Mathematics 1010 online

An equation involving radicals is called a radical equation (naturally). To solve it you simply apply our general principle:

To solve an equation figure out what bothers you and then do the same thing on both sides of the equation to get rid of it.

To get rid of a radical you take it to a power that will change the rational exponent to a natural number. This will work if the radical is on one side of the equation by itself.

Let's look at a few simple examples:

Suppose

We proceed as follows:

Here is a slightly more complicated problem:

We obtain

Our last example shows how to get rid of more than one radical:

To get rid of the square roots we isolate them and square one at a time:

In each case, we check our answer by substituting it in the original equation. For example, in the last equation we obtain:

Later in the course we will consider more complicated cases of radical equations.

Numerical Values

The radicals in the above examples were all natural numbers. This is due only to a judicious choice of examples. Frequently the roots occurring in applications are irrational numbers with decimal expansions that never repeat or terminate. The following table lists approximations of a few specific radicals.

Radical expressions yield roots and are the inverse of exponential expressions.

Learning Objectives

Describe the root of a number in terms of exponentiation

Key Takeaways

Key Points

• Roots are the inverse operation of exponentiation. This means that if $\sqrt [ n ]{ x } = r$, then ${r}^{n}=x$.
• The square root of a value is the number that when squared results in the initial value. In other words, $\sqrt{y}=x$ if $x^2=y$.
• The cube root of a value is the number that when cubed results in the initial value. In other words, $\sqrt[3]{y} = x$ if $x^3 = y$.

Key Terms

• root: A number that when raised to a specified power yields a specified number or expression.
• radical expression: A mathematical expression that contains a root, written in the form $\sqrt[n]{a}$.
• cube root: A root of degree 3, written in the form $\sqrt[3]{a}$.
• square root: A root of degree 2, written in the form $\sqrt{a}$.

Roots are the inverse operation of exponentiation. Mathematical expressions with roots are called radical expressions and can be easily recognized because they contain a radical symbol ($\sqrt{}$).

Recall that exponents signify that we should multiply a given integer a certain number of times. For example, $7^2$ tells us that we should multiply 7 by itself two times:

$7^2 = 7 \cdot 7 = 49$

Since roots are the inverse operation of exponentiation, they allow us to work backwards from the solution of an exponential expression to the number in the base of the expression.

For example, the following is a radical expression that reverses the above solution, working backwards from 49 to its square root:

$\sqrt{49} = 7$

In this expression, the symbol is known as the “radical,” and the solution of 7 is called the “root.”

Finding the value for a particular root can be much more difficult than solving an exponential expression. For now, it is important simplify to recognize the relationship between roots and exponents: if a root $r$ is defined as the $n \text{th}$ root of $x$, it is represented as

$\sqrt [ n ]{ x } = r$

Because roots are the inverse of exponents, we can cancel out the root in this equation by raising the answer to the nth power:

$\left( \sqrt [ n ]{ x }\right) ^n = \left(r\right) ^n$

To simplify:

${r}^{n}=x$

Square Roots

If the square root of a number $x$ is calculated, the result is a number that when squared (i.e., when raised to an exponent of 2) gives the original number $x$. This can be written symbolically as follows: $\sqrt x = y$ if ${y}^{2}=x$. This rule applies to the series of real numbers ${ y }^{ 2 }\ge 0$, regardless of the value of $y$. As such, when $x<0$ then $\sqrt x$ cannot be defined.

For example, consider the following: $\sqrt{36}$. This is read as “the square root of 36” or “radical 36.” You may recognize that $6^2 = 6 \cdot 6 = 36$, and therefore conclude that 6 is the root of $\sqrt{36}$. Thus we have the answer, $\sqrt{36} = 6$.

Cube Roots

The cube root of a number ($\sqrt [ 3 ]{x}$) can also be calculated. The cube root of a value $x$ is the number that when cubed (i.e., when raised to an exponent of 3) yields the original number $x$.

For example, the cube root of 8 is 2 because $2^3 = 2 \cdot 2\cdot 2=8$. This can also be written as $\sqrt[3]{8}=2$.

Other Roots

There are an infinite number of possible roots all in the form of $\sqrt [n]{a}$. Any non-zero integer can be substituted for $n$. For example, $\sqrt[4]{a}$ is called the “fourth root of $a$,” and $\sqrt[20]{a}$ is called the “twentieth root of $a$.”

Note that for any such root, if $\sqrt [n]{a} = b$ then ${b}^{n} = a$. As an example, consider $\sqrt[4]{2401} = 7$. $7^4 = 7\cdot 7\cdot 7\cdot 7 = 2401$.

Radicals and exponents have particular requirements for addition and subtraction while multiplication is carried out more freely.

Learning Objectives

Differentiate between correct and incorrect uses of operations on radical expressions

Key Takeaways

Key Points

• Subtraction follows the same rules as addition: the radicand must be the same.
• Multiplication of radicals simply requires that we multiply the term under the radical signs.

Key Terms

• radicand: The number or expression whose square root or other root is being considered; e.g., the 3 in $\sqrt[n]{3}$. More simply, the number under the radical.
• radical expression: An expression that represents the root of a number or quantity.

Roots are the inverse operation for exponents. An expression with roots is called a radical expression. It’s easy, although perhaps tedious, to compute exponents given a root. For instance $7\cdot7\cdot7\cdot7 = 49\cdot49 = 2401$. So, we know the fourth root of 2401 is 7, and the square root of 2401 is 49. What is the third root of 2401? Finding the value for a particular root is difficult. This is because exponentiation is a different kind of function than addition, subtraction, multiplication, and division.

Let’s go through some basic mathematical operations with radicals and exponents.

$a\sqrt{b}+c\sqrt{b} = (a+c)\sqrt{b}$

Let’s plug some numbers in place of the variables:

$\sqrt 3 +2\sqrt 3 = 3\sqrt 3$

Subtraction follows the same rules as addition:

$a\sqrt b - c\sqrt b = (a-c)\sqrt b$

For example:

$3\sqrt 3 -2\sqrt 3 = \sqrt 3$

Multiplication of radicals simply requires that we multiply the variable under the radical signs.

$\sqrt a \cdot \sqrt b = \sqrt {a\cdot b}$

Some examples with real numbers:

$\sqrt 3 \cdot \sqrt 6 = \sqrt {18}$

This equation can actually be simplified further; we will go over simplification in another section.

A radical expression can be simplified if:

1. the value under the radical sign can be written as an exponent,
2. there are fractions under the radical sign,
3. there is a radical expression in the denominator.

For example, the radical expression $\displaystyle \sqrt{\frac{16}{3}}$ can be simplified by first removing the squared value from the numerator.

$\displaystyle \sqrt{\frac{16}{3}} = \sqrt{\frac{4^2}{3}} = 4\sqrt{\frac{1}{3}}$

Then, the fraction under the radical sign can be addressed, and the radical in the numerator can again be simplified.

$\displaystyle 4\sqrt{\frac{1}{3}} = \frac{4\sqrt{1}}{\sqrt{3}} = \frac{4}{\sqrt{3}}$

Finally, the radical needs to be removed from the denominator.

$\displaystyle \frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3} = \frac{4}{3}\sqrt{3}$

Root rationalization is a process by which any roots in the denominator of an irrational fraction are eliminated.

Learning Objectives

Convert between fractions with and without rationalized denominators

Key Takeaways

Key Points

• To rationalize the denominator, multiply both the numerator and denominator by the radical in the denominator.

Key Terms

• rationalization: A process by which radicals in the denominator of an fraction are eliminated.

In mathematics, we are often given terms in the form of fractions with radicals in the numerator and/or denominator. When we are given expressions that involve radicals in the denominator, it makes it easier to evaluate the expression if we rewrite it in a way that the radical is no longer in the denominator. This process is called rationalizing the denominator.

Before we begin, remember that whatever we do to one side of an algebraic equation, we must also do to the other side. This same principle can be applied to fractions: whatever we do to the numerator, we must also do to the denominator, and vice versa.

Let’s look at an example to illustrate the process of rationalizing the denominator.

You are given the fraction $\frac{10}{\sqrt{3}}$, and you want to simplify it by eliminating the radical from the denominator. Recall that a radical multiplied by itself equals its radicand, or the value under the radical sign. Therefore, multiply the top and bottom of the fraction by $\frac{\sqrt{3}}{\sqrt{3}}$, and watch how the radical expression disappears from the denominator:$\displaystyle \frac{10}{\sqrt{3}} \cdot\frac{\sqrt{3}}{\sqrt{3}} = {\frac{10\cdot\sqrt{3}}{{\sqrt{3}}^2}} = {\frac{10\sqrt{3}}{3}}$

Imaginary Numbers

There is no such value such that when squared it results in a negative value; we therefore classify roots of negative numbers as “imaginary.”

Learning Objectives

Explain what imaginary numbers are and why they are needed in mathematics

Key Takeaways

Key Points

• There is no such value such that when squared it results in a negative value.More specifically, solving $x^2=-1$ for $x$ results in a “number” that would not be a real number, referred to as an imaginary number.
• The imaginary number, $i$, is defined as the square root of -1: $i=\sqrt{-1}$.

Key Terms

• imaginary number: The square root of -1.

A radical expression represents the root of a given quantity. What does it mean, then, if the value under the radical is negative, such as in $\displaystyle \sqrt{-1}$? There is no real value such that when multiplied by itself it results in a negative value. This means that there is no real value of $x$ that would make $x^2 =-1$ a true statement.

That is where imaginary numbers come in. When the radicand (the value under the radical sign) is negative, the root of that value is said to be an imaginary number. Specifically, the imaginary number, $i$, is defined as the square root of -1: thus, $i=\sqrt{-1}$.

We can write the square root of any negative number in terms of $i$. Here are some examples:

• $\sqrt{-25}=\sqrt{25\cdot-1}=\sqrt{25}\cdot\sqrt{-1}=5i$
• $\sqrt{-18} = \sqrt{2\cdot9\cdot-1} = \sqrt{2} \cdot \sqrt{9} \cdot \sqrt{-1} = 3i\sqrt{2}$

The properties of exponents, which we've talked about earlier, tell us among other things that

$$\begin{pmatrix} xy \end{pmatrix}^{a}=x^{a}y^{a}$$

$$\begin{pmatrix} \frac{x}{y} \end{pmatrix}^{a}=\frac{x^{a}}{y^{a}}$$

We also know that

$$\sqrt[a]{x}=x^{\frac{1}{a}}$$

$$or$$

$$\sqrt{x}=x^{\frac{1}{2}}$$

If we combine these two things then we get the product property of radicals and the quotient property of radicals. These two properties tell us that the square root of a product equals the product of the square roots of the factors.

$$\sqrt{xy}=\sqrt{x}\cdot \sqrt{y}$$

$$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}$$

$$where\:\: x\geq 0,y\geq 0$$

The answer can't be negative and x and y can't be negative since we then wouldn't get a real answer. In the same way we know that

$$\sqrt{x^{2}}=x\: \: where\: \: x\geq 0$$

These properties can be used to simplify radical expressions. A radical expression is said to be in its simplest form if there are

no perfect square factors other than 1 in the radicand

$$\sqrt{16x}=\sqrt{16}\cdot \sqrt{x}=\sqrt{4^{2}}\cdot \sqrt{x}=4\sqrt{x}$$

no fractions in the radicand and

$$\sqrt{\frac{25}{16}x^{2}}=\frac{\sqrt{25}}{\sqrt{16}}\cdot \sqrt{x^{2}}=\frac{5}{4}x$$

no radicals appear in the denominator of a fraction.

$$\sqrt{\frac{15}{16}}=\frac{\sqrt{15}}{\sqrt{16}}=\frac{\sqrt{15}}{4}$$

If the denominator is not a perfect square you can rationalize the denominator by multiplying the expression by an appropriate form of 1 e.g.

$$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\cdot {\color{green} {\frac{\sqrt{y}}{\sqrt{y}}}}=\frac{\sqrt{xy}}{\sqrt{y^{2}}}=\frac{\sqrt{xy}}{y}$$

Binomials like

$$x\sqrt{y}+z\sqrt{w}\: \: and\: \: x\sqrt{y}-z\sqrt{w}$$

are called conjugates to each other. The product of two conjugates is always a rational number which means that you can use conjugates to rationalize the denominator e.g.

$$\frac{x}{4+\sqrt{x}}=\frac{x\left ( {\color{green} {4-\sqrt{x}}} \right )}{\left ( 4+\sqrt{x} \right )\left ( {\color{green}{ 4-\sqrt{x}}} \right )}=$$

$$=\frac{x\left ( 4-\sqrt{x} \right )}{16-\left ( \sqrt{x} \right )^{2}}=\frac{4x-x\sqrt{x}}{16-x}$$

Video lesson

$$\frac{x}{5-\sqrt{x}}$$

Math Skills Overview Guide

Definitions:

From Wolfram MathWorld:

Exponents - An exponent is the power p in an expression of the form $$a^p$$ The process of performing the operation of raising a base to a given power is known as exponentiation. (where a ≠0)

Radicals - The symbol $$\sqrt[n]{x}$$ used to indicate a root is called a radical and is therefore read "x radical n," or "the nth root of x." In the radical symbol, the horizontal line is called the vinculum, the quantity under the vinculum is called the radicand, and the quantity n written to the left is called the index.

The special case $$\sqrt[2]{x}$$ is written $$\sqrt{x}$$ and is called the square root of x. $$\sqrt[3]{x}$$ is called the cube root.

MORE:

Radical - The √ symbol that is used to denote square root or nth roots.

Radical Expression - A radical expression is an expression containing a square root.

Sours: https://davenport.libguides.com/math-skills-overview/exponents/definition
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We’ll open this section with the definition of the radical. If $$n$$ is a positive integer that is greater than 1 and $$a$$ is a real number then,

$\sqrt[n]{a} = {a^{\frac{1}{n}}}$

where $$n$$ is called the index, $$a$$ is called the radicand, and the symbol $$\sqrt {}$$ is called the radical. The left side of this equation is often called the radical form and the right side is often called the exponent form.

From this definition we can see that a radical is simply another notation for the first rational exponent that we looked at in the rational exponents section.

Note as well that the index is required in these to make sure that we correctly evaluate the radical. There is one exception to this rule and that is square root. For square roots we have,

$\sqrt[2]{a} = \sqrt a$

In other words, for square roots we typically drop the index.

Let’s do a couple of examples to familiarize us with this new notation.

Example 1Write each of the following radicals in exponent form.
1. $$\sqrt[4]{{16}}$$
2. $$\sqrt[{10}]{{8x}}$$
3. $$\sqrt {{x^2} + {y^2}}$$
Show Solution

Intro text prior to part solutions.

a $$\sqrt[4]{{16}} = {16^{\frac{1}{4}}}$$

b $$\sqrt[{10}]{{8x}} = {\left( {8x} \right)^{\frac{1}{{10}}}}$$

c $$\sqrt {{x^2} + {y^2}} = {\left( {{x^2} + {y^2}} \right)^{\frac{1}{2}}}$$

As seen in the last two parts of this example we need to be careful with parenthesis. When we convert to exponent form and the radicand consists of more than one term then we need to enclose the whole radicand in parenthesis as we did with these two parts. To see why this is consider the following,

$8{x^{\frac{1}{{10}}}}$

From our discussion of exponents in the previous sections we know that only the term immediately to the left of the exponent actually gets the exponent. Therefore, the radical form of this is,

$8{x^{\frac{1}{{10}}}} = 8\,\,\sqrt[{10}]{x} \ne \sqrt[{10}]{{8x}}$

So, we once again see that parenthesis are very important in this class. Be careful with them.

Since we know how to evaluate rational exponents we also know how to evaluate radicals as the following set of examples shows.

Example 2Evaluate each of the following.
1. $$\sqrt {16}$$ and $$\sqrt[4]{{16}}$$
2. $$\sqrt[5]{{243}}$$
3. $$\sqrt[4]{{1296}}$$
4. $$\sqrt[3]{{ - 125}}$$
5. $$\sqrt[4]{{ - 16}}$$
Show All Solutions Hide All Solutions
Show Discussion

To evaluate these we will first convert them to exponent form and then evaluate that since we already know how to do that.

a$$\sqrt {16}$$ and $$\sqrt[4]{{16}}$$ Show Solution

These are together to make a point about the importance of the index in this notation. Let’s take a look at both of these.

$\sqrt {16} = {16^{\frac{1}{2}}} = 4\hspace{0.25in}\hspace{0.25in}{\mbox{because }}{4^2} = 16$ $\sqrt[4]{{16}} = {16^{\frac{1}{4}}} = 2\hspace{0.25in}\hspace{0.25in}{\mbox{because }}{{\mbox{2}}^{\mbox{4}}} = 16$

So, the index is important. Different indexes will give different evaluations so make sure that you don’t drop the index unless it is a 2 (and hence we’re using square roots).

b$$\sqrt[5]{{243}}$$ Show Solution

$\sqrt[5]{{243}} = {243^{\frac{1}{5}}} = 3\hspace{0.25in}\hspace{0.25in}{\mbox{because }}{{\mbox{3}}^{\mbox{5}}} = 243$

c$$\sqrt[4]{{1296}}$$ Show Solution

$\sqrt[4]{{1296}} = {1296^{\frac{1}{4}}} = 6\hspace{0.25in}\hspace{0.25in}{\mbox{because }}{{\mbox{6}}^{\mbox{4}}} = 1296$

d$$\sqrt[3]{{ - 125}}$$ Show Solution

$\sqrt[3]{{ - 125}} = {\left( { - 125} \right)^{\frac{1}{3}}} = - 5\hspace{0.25in}{\mbox{because }}{\left( { - 5} \right)^{\mbox{3}}} = - 125$

e$$\sqrt[4]{{ - 16}}$$ Show Solution
$\sqrt[4]{{ - 16}} = {\left( { - 16} \right)^{\frac{1}{4}}}$

As we saw in the integer exponent section this does not have a real answer and so we can’t evaluate the radical of a negative number if the index is even. Note however that we can evaluate the radical of a negative number if the index is odd as the previous part shows.

Let’s briefly discuss the answer to the first part in the above example. In this part we made the claim that $$\sqrt {16} = 4$$ because $${4^2} = 16$$. However, 4 isn’t the only number that we can square to get 16. We also have $${\left( { - 4} \right)^2} = 16$$. So, why didn’t we use -4 instead? There is a general rule about evaluating square roots (or more generally radicals with even indexes). When evaluating square roots we ALWAYS take the positive answer. If we want the negative answer we will do the following.

$- \sqrt {16} = - 4$

This may not seem to be all that important, but in later topics this can be very important. Following this convention means that we will always get predictable values when evaluating roots.

Note that we don’t have a similar rule for radicals with odd indexes such as the cube root in part (d) above. This is because there will never be more than one possible answer for a radical with an odd index.

We can also write the general rational exponent in terms of radicals as follows.

${a^{\frac{m}{n}}} = {\left( {{a^{\frac{1}{n}}}} \right)^m} = {\left( {\sqrt[n]{a}} \right)^m}\hspace{0.25in}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\hspace{0.25in}{a^{\frac{m}{n}}} = {\left( {{a^m}} \right)^{\frac{1}{n}}} = \sqrt[n]{{{a^m}}}$

Properties

If $$n$$ is a positive integer greater than 1 and both $$a$$ and $$b$$ are positive real numbers then,

1. $$\sqrt[n]{{{a^n}}} = a$$
2. $$\sqrt[n]{{ab}} = \sqrt[n]{a}\,\sqrt[n]{b}$$
3. $$\displaystyle \sqrt[n]{{\frac{a}{b}}} = \frac{{\sqrt[n]{a}}}{{\sqrt[n]{b}}}$$

Note that on occasion we can allow $$a$$ or $$b$$ to be negative and still have these properties work. When we run across those situations we will acknowledge them. However, for the remainder of this section we will assume that $$a$$ and $$b$$ must be positive.

Also note that while we can “break up” products and quotients under a radical we can’t do the same thing for sums or differences. In other words,

$\sqrt[n]{{a + b}} \ne \sqrt[n]{a} + \sqrt[n]{b}\hspace{0.25in}\hspace{0.25in}{\mbox{AND}}\hspace{0.25in}\hspace{0.25in}\sqrt[n]{{a - b}} \ne \sqrt[n]{a} - \sqrt[n]{b}$

If you aren’t sure that you believe this consider the following quick number example.

$5 = \sqrt {25} = \sqrt {9 + 16} \ne \sqrt 9 + \sqrt {16} = 3 + 4 = 7$

If we “break up” the root into the sum of the two pieces we clearly get different answers! So, be careful to not make this very common mistake!

We are going to be simplifying radicals shortly so we should next define simplified radical form. A radical is said to be in simplified radical form (or just simplified form) if each of the following are true.

1. All exponents in the radicand must be less than the index.
2. Any exponents in the radicand can have no factors in common with the index.
3. No fractions appear under a radical.
4. No radicals appear in the denominator of a fraction.

In our first set of simplification examples we will only look at the first two. We will need to do a little more work before we can deal with the last two.

Example 3Simplify each of the following. Assume that $$x$$, $$y$$, and $$z$$ are positive.
1. $$\sqrt {{y^7}}$$
2. $$\sqrt[9]{{{x^6}}}$$
3. $$\sqrt {18{x^6}{y^{11}}}$$
4. $$\sqrt[4]{{32{x^9}{y^5}{z^{12}}}}$$
5. $$\sqrt[5]{{{x^{12}}{y^4}{z^{24}}}}$$
6. $$\sqrt[3]{{9{x^2}}}\,\sqrt[3]{{6{x^2}}}$$
Show All Solutions Hide All Solutions
a$$\sqrt {{y^7}}$$ Show Solution

In this case the exponent (7) is larger than the index (2) and so the first rule for simplification is violated. To fix this we will use the first and second properties of radicals above. So, let’s note that we can write the radicand as follows.

${y^7} = {y^6}y = {\left( {{y^3}} \right)^2}y$

So, we’ve got the radicand written as a perfect square times a term whose exponent is smaller than the index. The radical then becomes,

$\sqrt {{y^7}} = \sqrt {{{\left( {{y^3}} \right)}^2}y}$

Now use the second property of radicals to break up the radical and then use the first property of radicals on the first term.

$\sqrt {{y^7}} = \sqrt {{{\left( {{y^3}} \right)}^2}} \,\,\sqrt y = {y^3}\sqrt y$

This now satisfies the rules for simplification and so we are done.

Before moving on let’s briefly discuss how we figured out how to break up the exponent as we did. To do this we noted that the index was 2. We then determined the largest multiple of 2 that is less than 7, the exponent on the radicand. This is 6. Next, we noticed that 7=6+1.

Finally, remembering several rules of exponents we can rewrite the radicand as,

${y^7} = {y^6}y = {y^{\left( 3 \right)\left( 2 \right)}}y = {\left( {{y^3}} \right)^2}y$

In the remaining examples we will typically jump straight to the final form of this and leave the details to you to check.

b$$\sqrt[9]{{{x^6}}}$$ Show Solution

This radical violates the second simplification rule since both the index and the exponent have a common factor of 3. To fix this all we need to do is convert the radical to exponent form do some simplification and then convert back to radical form.

$\sqrt[9]{{{x^6}}} = {\left( {{x^6}} \right)^{\frac{1}{9}}} = {x^{\frac{6}{9}}} = {x^{\frac{2}{3}}} = {\left( {{x^2}} \right)^{\frac{1}{3}}} = \sqrt[3]{{{x^2}}}$

c$$\sqrt {18{x^6}{y^{11}}}$$ Show Solution

Now that we’ve got a couple of basic problems out of the way let’s work some harder ones. Although, with that said, this one is really nothing more than an extension of the first example.

There is more than one term here but everything works in exactly the same fashion. We will break the radicand up into perfect squares times terms whose exponents are less than 2 (i.e. 1).

$18{x^6}{y^{11}} = 9{x^6}{y^{10}}\left( {2y} \right) = 9{\left( {{x^3}} \right)^2}{\left( {{y^5}} \right)^2}\left( {2y} \right)$

Don’t forget to look for perfect squares in the number as well.

Now, go back to the radical and then use the second and first property of radicals as we did in the first example.

$\sqrt {18{x^6}{y^{11}}} = \sqrt {9{{\left( {{x^3}} \right)}^2}{{\left( {{y^5}} \right)}^2}\left( {2y} \right)} = \sqrt 9 \sqrt {{{\left( {{x^3}} \right)}^2}} \sqrt {{{\left( {{y^5}} \right)}^2}} \sqrt {2y} = 3{x^3}{y^5}\sqrt {2y}$

Note that we used the fact that the second property can be expanded out to as many terms as we have in the product under the radical. Also, don’t get excited that there are no $$x$$’s under the radical in the final answer. This will happen on occasion.

d$$\sqrt[4]{{32{x^9}{y^5}{z^{12}}}}$$ Show Solution

This one is similar to the previous part except the index is now a 4. So, instead of get perfect squares we want powers of 4. This time we will combine the work in the previous part into one step.

$\sqrt[4]{{32{x^9}{y^5}{z^{12}}}} = \sqrt[4]{{16{x^8}{y^4}{z^{12}}\left( {2xy} \right)}} = \sqrt[4]{{16}}\sqrt[4]{{{{\left( {{x^2}} \right)}^4}}}\sqrt[4]{{{y^4}}}\sqrt[4]{{{{\left( {{z^3}} \right)}^4}}}\sqrt[4]{{2xy}} = 2{x^2}y\,{z^3}\sqrt[4]{{2xy}}$

e$$\sqrt[5]{{{x^{12}}{y^4}{z^{24}}}}$$ Show Solution

Again this one is similar to the previous two parts.

$\sqrt[5]{{{x^{12}}{y^4}{z^{24}}}} = \sqrt[5]{{{x^{10}}{z^{20}}\left( {{x^2}{y^4}{z^4}} \right)}} = \sqrt[5]{{{{\left( {{x^2}} \right)}^5}}}\sqrt[5]{{{{\left( {{z^4}} \right)}^5}}}\sqrt[5]{{{x^2}{y^4}{z^4}}} = {x^2}{z^4}\sqrt[5]{{{x^2}{y^4}{z^4}}}$

In this case don’t get excited about the fact that all the $$y$$’s stayed under the radical. That will happen on occasion.

f$$\sqrt[3]{{9{x^2}}}\,\sqrt[3]{{6{x^2}}}$$ Show Solution

This last part seems a little tricky. Individually both of the radicals are in simplified form. However, there is often an unspoken rule for simplification. The unspoken rule is that we should have as few radicals in the problem as possible. In this case that means that we can use the second property of radicals to combine the two radicals into one radical and then we’ll see if there is any simplification that needs to be done.

$\sqrt[3]{{9{x^2}}}\,\sqrt[3]{{6{x^2}}} = \sqrt[3]{{\left( {9{x^2}} \right)\left( {6{x^2}} \right)}}\, = \sqrt[3]{{54{x^4}}}$

Now that it’s in this form we can do some simplification.

$\sqrt[3]{{9{x^2}}}\,\sqrt[3]{{6{x^2}}} = \sqrt[3]{{27{x^3}\left( {2x} \right)}} = \sqrt[3]{{27{x^3}}}\sqrt[3]{{2x}} = 3x\sqrt[3]{{2x}}$

Before moving into a set of examples illustrating the last two simplification rules we need to talk briefly about adding/subtracting/multiplying radicals. Performing these operations with radicals is much the same as performing these operations with polynomials. If you don’t remember how to add/subtract/multiply polynomials we will give a quick reminder here and then give a more in depth set of examples the next section.

Recall that to add/subtract terms with $$x$$ in them all we need to do is add/subtract the coefficients of the $$x$$. For example,

$4x + 9x = \left( {4 + 9} \right)x = 13x\hspace{0.25in}\hspace{0.25in}3x - 11x = \left( {3 - 11} \right)x = - 8x$

$4\sqrt x + 9\sqrt x = \left( {4 + 9} \right)\sqrt x = 13\sqrt x \hspace{0.25in}3\,\,\sqrt[{10}]{5} - 11\,\,\sqrt[{10}]{5} = \left( {3 - 11} \right)\sqrt[{10}]{5} = - 8\,\,\sqrt[{10}]{5}$

We’ve already seen some multiplication of radicals in the last part of the previous example. If we are looking at the product of two radicals with the same index then all we need to do is use the second property of radicals to combine them then simplify. What we need to look at now are problems like the following set of examples.

Example 4Multiply each of the following. Assume that $$x$$ is positive.
1. $$\left( {\sqrt x + 2} \right)\left( {\sqrt x - 5} \right)$$
2. $$\left( {3\,\sqrt x - \sqrt y } \right)\left( {2\sqrt x - 5\sqrt y } \right)$$
3. $$\left( {5\sqrt x + 2} \right)\left( {5\sqrt x - 2} \right)$$
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In all of these problems all we need to do is recall how to FOIL binomials. Recall,

$\left( {3x - 5} \right)\left( {x + 2} \right) = 3x\left( x \right) + 3x\left( 2 \right) - 5\left( x \right) - 5\left( 2 \right) = 3{x^2} + 6x - 5x - 10 = 3{x^2} + x - 10$

With radicals we multiply in exactly the same manner. The main difference is that on occasion we’ll need to do some simplification after doing the multiplication

a$$\left( {\sqrt x + 2} \right)\left( {\sqrt x - 5} \right)$$ Show Solution
\begin{align*}\left( {\sqrt x + 2} \right)\left( {\sqrt x - 5} \right) & = \sqrt x \left( {\sqrt x } \right) - 5\sqrt x + 2\sqrt x - 10\\ & = \sqrt {{x^2}} - 3\sqrt x - 10\\ & = x - 3\sqrt x - 10\end{align*}

As noted above we did need to do a little simplification on the first term after doing the multiplication.

b$$\left( {3\,\sqrt x - \sqrt y } \right)\left( {2\sqrt x - 5\sqrt y } \right)$$ Show Solution

Don’t get excited about the fact that there are two variables here. It works the same way!

\begin{align*}\left( {3\,\sqrt x - \sqrt y } \right)\left( {2\sqrt x - 5\sqrt y } \right) & = 6\sqrt {{x^2}} - 15\sqrt x \,\sqrt y - 2\sqrt x \,\sqrt y + 5\sqrt {{y^2}} \\ & = 6x - 15\sqrt {xy} - 2\sqrt {xy} + 5y\\ & = 6x - 17\sqrt {xy} + 5y\end{align*}

Again, notice that we combined up the terms with two radicals in them.

c$$\left( {5\sqrt x + 2} \right)\left( {5\sqrt x - 2} \right)$$ Show Solution

Not much to do with this one.

$\left( {5\sqrt x + 2} \right)\left( {5\sqrt x - 2} \right) = 25\sqrt {{x^2}} - 10\sqrt x + 10\sqrt x - 4 = 25x - 4$

Notice that, in this case, the answer has no radicals. That will happen on occasion so don’t get excited about it when it happens.

The last part of the previous example really used the fact that

$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

If you don’t recall this formula we will look at it in a little more detail in the next section.

Okay, we are now ready to take a look at some simplification examples illustrating the final two rules. Note as well that the fourth rule says that we shouldn’t have any radicals in the denominator. To get rid of them we will use some of the multiplication ideas that we looked at above and the process of getting rid of the radicals in the denominator is called rationalizing the denominator. In fact, that is really what this next set of examples is about. They are really more examples of rationalizing the denominator rather than simplification examples.

Example 5Rationalize the denominator for each of the following. Assume that $$x$$ is positive.
1. $$\displaystyle \frac{4}{{\sqrt x }}$$
2. $$\displaystyle \sqrt[5]{{\frac{2}{{{x^3}}}}}$$
3. $$\displaystyle \frac{1}{{3 - \sqrt x }}$$
4. $$\displaystyle \frac{5}{{4\sqrt x + \sqrt 3 }}$$
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There are really two different types of problems that we’ll be seeing here. The first two parts illustrate the first type of problem and the final two parts illustrate the second type of problem. Both types are worked differently.

a$$\displaystyle \frac{4}{{\sqrt x }}$$ Show Solution

In this case we are going to make use of the fact that $$\sqrt[n]{{{a^n}}} = a$$. We need to determine what to multiply the denominator by so that this will show up in the denominator. Once we figure this out we will multiply the numerator and denominator by this term.

Here is the work for this part.

$\frac{4}{{\sqrt x }} = \frac{4}{{\sqrt x }}\frac{{\sqrt x }}{{\sqrt x }} = \frac{{4\sqrt x }}{{\sqrt {{x^2}} }} = \frac{{4\sqrt x }}{x}$

Remember that if we multiply the denominator by a term we must also multiply the numerator by the same term. In this way we are really multiplying the term by 1 (since $$\frac{a}{a} = 1$$) and so aren’t changing its value in any way.

b$$\displaystyle \sqrt[5]{{\frac{2}{{{x^3}}}}}$$ Show Solution

We’ll need to start this one off with first using the third property of radicals to eliminate the fraction from underneath the radical as is required for simplification.

$\sqrt[5]{{\frac{2}{{{x^3}}}}} = \frac{{\sqrt[5]{2}}}{{\sqrt[5]{{{x^3}}}}}$

Now, in order to get rid of the radical in the denominator we need the exponent on the x to be a 5. This means that we need to multiply by $$\sqrt[5]{{{x^2}}}$$ so let’s do that.

$\sqrt[5]{{\frac{2}{{{x^3}}}}} = \frac{{\sqrt[5]{2}}}{{\sqrt[5]{{{x^3}}}}}\frac{{\sqrt[5]{{{x^2}}}}}{{\sqrt[5]{{{x^2}}}}} = \frac{{\sqrt[5]{{2{x^2}}}}}{{\sqrt[5]{{{x^5}}}}} = \frac{{\sqrt[5]{{2{x^2}}}}}{x}$

c$$\displaystyle \frac{1}{{3 - \sqrt x }}$$ Show Solution

In this case we can’t do the same thing that we did in the previous two parts. To do this one we will need to instead to make use of the fact that

$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

When the denominator consists of two terms with at least one of the terms involving a radical we will do the following to get rid of the radical.

$\frac{1}{{3 - \sqrt x }} = \frac{1}{{\left( {3 - \sqrt x } \right)}}\frac{{3 + \sqrt x }}{{\left( {3 + \sqrt x } \right)}} = \frac{{3 + \sqrt x }}{{\left( {3 - \sqrt x } \right)\left( {3 + \sqrt x } \right)}} = \frac{{3 + \sqrt x }}{{9 - x}}$

So, we took the original denominator and changed the sign on the second term and multiplied the numerator and denominator by this new term. By doing this we were able to eliminate the radical in the denominator when we then multiplied out.

d$$\displaystyle \frac{5}{{4\sqrt x + \sqrt 3 }}$$ Show Solution

This one works exactly the same as the previous example. The only difference is that both terms in the denominator now have radicals. The process is the same however.

$\frac{5}{{4\sqrt x + \sqrt 3 }} = \frac{5}{{\left( {4\sqrt x + \sqrt 3 } \right)}}\frac{{\left( {4\sqrt x - \sqrt 3 } \right)}}{{\left( {4\sqrt x - \sqrt 3 } \right)}} = \frac{{5\left( {4\sqrt x - \sqrt 3 } \right)}}{{\left( {4\sqrt x + \sqrt 3 } \right)\left( {4\sqrt x - \sqrt 3 } \right)}} = \frac{{5\left( {4\sqrt x - \sqrt 3 } \right)}}{{16x - 3}}$

Rationalizing the denominator may seem to have no real uses and to be honest we won’t see many uses in an Algebra class. However, if you are on a track that will take you into a Calculus class you will find that rationalizing is useful on occasion at that level.

We will close out this section with a more general version of the first property of radicals. Recall that when we first wrote down the properties of radicals we required that $$a$$ be a positive number. This was done to make the work in this section a little easier. However, with the first property that doesn’t necessarily need to be the case.

Here is the property for a general $$a$$ (i.e. positive or negative)

$\sqrt[n]{{{a^n}}} = \left\{ {\begin{array}{*{20}{l}}{\left| a \right|}&{{\mbox{if }}n{\mbox{ is even}}}\\a&{{\mbox{if }}n{\mbox{ is odd}}}\end{array}} \right.$

where $$\left| a \right|$$ is the absolute value of $$a$$. If you don’t recall absolute value we will cover that in detail in a section in the next chapter. All that you need to do is know at this point is that absolute value always makes $$a$$ a positive number.

So, as a quick example this means that,

$\sqrt[8]{{{x^8}}} = \left| x \right|\hspace{0.25in}\hspace{0.25in}{\mbox{AND}}\hspace{0.25in}\hspace{0.25in}\,\,\,\,\,\,\sqrt[{11}]{{{x^{11}}}} = x$

For square roots this is,

$\sqrt {{x^2}} = \left| x \right|$

This will not be something we need to worry all that much about here, but again there are topics in courses after an Algebra course for which this is an important idea so we needed to at least acknowledge it.

Purplemath

"Roots" (or "radicals") are the "opposite" operation of applying exponents; we can "undo" a power with a radical, and we can "undo" a radical with a power. For instance, if we square 2, we get 4, and if we "take the square root of 4", we get 2; if we square 3, we get 9, and if we "take the square root of 9", we get 3. In mathematical notation, the previous sentence means the following:

The "  " symbol used above is called the "radical"symbol. (Technically, just the "check mark" part of the symbol is the radical; the line across the top is called the "vinculum".) The expression "  " is read as "root nine", "radical nine", or "the square root of nine".

We can raise numbers to powers other than just 2; we can cube things (being raising things to the third power, or "to the power 3"), raise them to the fourth power (or "to the power 4"), raise them to the 100th power, and so forth. In the same way, we can take the cube root of a number, the fourth root, the 100th root, and so forth. Just as the square root undoes squaring, so also the cube root undoes cubing, the fourth root undoes raising things to the fourth power, et cetera. To indicate some root other than a square root when writing, we use the same radical symbol as for the square root, but we insert a number into the front of the radical, writing the number small and tucking it into the "check mark" part of the radical symbol. This tucked-in number corresponds to the root that you're taking. For instance, relating cubing and cube-rooting, we have:

The "3" in the radical above is called the "index" of the radical (the plural being "indices", pronounced "INN-duh-seez"); the "64" is "the argument of the radical", also called "the radicand". Perhaps because most of radicals you will see will be square roots, the index is not included on square roots. While "  " would be technically correct, I've never seen it used.

a square (second) root is written as:

a cube (third) root is written as:

a fourth root is written as:

a fifth root is written as:

We can take any counting number, square it, and end up with a nice neat number. But the process doesn't always work nicely when going backwards. For instance, consider , the square root of three. There is no nice neat number that squares to 3, so cannot be simplified as a nice whole number. We can deal with in either of two ways: If we are doing a word problem and are trying to find, say, the rate of speed, then we would grab our calculators and find the decimal approximation of :

Then we'd round the above value to an appropriate number of decimal places and use a real-world unit or label, like "1.7 ft/sec". On the other hand, we may be solving a plain old math exercise, something having no "practical" application. Then they would almost certainly want us to give the "exact" value, so we'd write our answer as being simply "".

Since most of what you'll be dealing with will be square roots (that is, second roots), most of this lesson will deal with them specifically.

Simplifying Square-Root Terms

To simplify a term containing a square root, we "take out" anything that is a "perfect square"; that is, we factor inside the radical symbol and then we take out in front of that symbol anything that has two copies of the same factor. For instance, 4 is the square of 2, so the square root of 4 contains two copies of the factor 2; thus, we can take a 2 out front, leaving nothing (but an understood 1) inside the radical, which we then drop:

Similarly, 49 is the square of 7, so it contains two copies of the factor 7:

And 225 is the square of 15, so it contains two copies of the factor 15, so:

Note that the value of the simplified radical is positive. While either of +2 and –2 might have been squared to get 4, "the square root of four" is defined to be only the positive option, +2. That is, the definition of the square root says that the square root will spit out only the positive root.

On a side note, let me emphasize that "evaluating" an expression (to find its one value) and "solving" an equation (to find its one or more, or no, solutions) are two very different things. In the first case, we're simplifying to find the one defined value for an expression. In the second case, we're looking for any and all values what will make the original equation true.

So, for instance, when we solve the equationx2 = 4, we are trying to find all possible values that might have been squared to get 4. But when we are just simplifying the expression , the ONLY answer is "2"; this positive result is called the "principal" root. (Other roots, such as –2, can be defined using graduate-school topics like "complex analysis" and "branch functions", but you won't need that for years, if ever.)

Oftentimes the argument of a radical is not a perfect square, but it may "contain" a square amongst its factors. To simplify this sort of radical, we need to factor the argument (that is, factor whatever is inside the radical symbol) and "take out" one copy of anything that is a square. That is, we find anything of which we've got a pair inside the radical, and we move one copy of it out front. When doing this, it can be helpful to use the fact that we can switch between the multiplication of roots and the root of a multiplication. In other words, we can use the fact that radicals can be manipulated similarly to powers:

There are various ways I can approach this simplification. One would be by factoring and then taking two different square roots. In particular, I'll start by factoring the argument, 144, into a product of squares:

Each of 9 and 16 is a square, so each of these can have its square root pulled out of the radical. The square root of 9 is 3 and the square root of 16 is 4. Then:

Then my solution is:

Another way to do the above simplification would be to remember our squares. You probably already knew that 122 = 144, so obviously the square root of 144 must be 12. But my steps above show how you can switch back and forth between the different formats (multiplication inside one radical, versus multiplication of two radicals) to help in the simplification process.

In case you're wondering, products of radicals are customarily written as shown above, using "multiplication by juxtaposition", meaning "they're put right next to one another, which we're using to mean that they're multiplied against each other". You could put a "times" symbol between the two radicals, but this isn't standard.

Neither of 24 and 6 is a square, but what happens if I multiply them inside one radical?

Now I do have something with squares in it, so I can simplify as before:

The argument of this radical, 75, factors as:

75 = 3 × 5 × 5

This factorization gives me two copies of the factor 5, but only one copy of the factor 3. Since I have two copies of 5, I can take 5 out front. Since I have only the one copy of 3, it'll have to stay behind in the radical. Then my answer is:

This answer is pronounced as "five, times root three", "five, times the square root of three", or, most commonly, just "five, root three".

When writing an expression containing radicals, it is proper form to put the radical at the end of the expression. Not only is "" non-standard, it is very hard to read, especially when hand-written. Is the 5 included in the square root, or not? (In our case here, it's not.)

And take care to write neatly, because "" is not the same as "". You don't want your handwriting to cause the reader to think you mean something other than what you'd intended.

You don't have to factor the radicand all the way down to prime numbers when simplifying. As soon as you see that you have a pair of factors or a perfect square, and that whatever remains will have nothing that can be pulled out of the radical, you've gone far enough.

Since 72 factors as 2×36, and since 36 is a perfect square, then:

Since there had been only one copy of the factor 2 in the factorization 2 × 6 × 6, the left-over 2 couldn't come out of the radical and had to be left behind.

The argument, 4,500, factors as:

45 × 100

= 5 × 9 × 100

I could continue factoring, but I know that 9 and 100 are squares, while 5 isn't, so I've gone as far as I need to. I'm ready to evaluate the square root:

Yes, I used "times" in my work above. No, you wouldn't include a "times" symbol in the final answer. I was using the "times" to help me keep things straight in my work. I used regular formatting for my hand-in answer. When doing your work, use whatever notation works well for you.

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