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Radicals and Rational Exponents

Evaluate square roots.

Evaluating Square Roots

When the square root of a number is squared, the result is the original number. Since \({4}^{2}=16,\) the square root of \(16\) is \(4.\) The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root.

In general terms, if \(a\) is a positive real number, then the square root of \(a\) is a number that, when multiplied by itself, gives \(a.\) The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals \(a.\) The square root obtained using a calculator is the principal square root.

The principal square root of \(a\) is written as \(\sqrt{a}.\) The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression.

Example

Try it.

Evaluate each expression.

  1. ⓐ \(\sqrt{100}\)
  2. ⓑ \(\sqrt{\sqrt{16}}\)
  3. ⓒ \(\sqrt{25+144}\)
  4. ⓓ \(\sqrt{49}-\sqrt{81}\)
Solution
  1. ⓐ \(\sqrt{100}=10\) because \({10}^{2}=100\)
  2. ⓑ \(\sqrt{\sqrt{16}}=\sqrt{4}=2\) because \({4}^{2}=16\) and \({2}^{2}=4\)
  3. ⓒ \(\sqrt{25+144}=\sqrt{169}=13\) because \({13}^{2}=169\)
  4. ⓓ \(\sqrt{49}-\sqrt{81}=7-9=-2\) because \({7}^{2}=49\) and \({9}^{2}=81\)

Using the Product Rule to Simplify Square Roots

To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite \(\sqrt{15}\) as \(\sqrt{3}⋅\sqrt{5}.\) We can also use the product rule to express the product of multiple radical expressions as a single radical expression.

Example

Try it.

Simplify the radical expression.

  1. ⓐ \(\sqrt{300}\)
  2. ⓑ \(\sqrt{162{a}^{5}{b}^{4}}\)
Solution
  1. ⓐ
    \(\begin{array}{ll}\sqrt{100⋅3} & \ \text{Factor perfect square from radicand}. \\ \sqrt{100}⋅\sqrt{3} & \ \text{Write radical expression as product of radical expressions}. \\ 10\sqrt{3} & \ \text{Simplify}.\end{array}\)
  2. ⓑ
    \(\begin{array}{ll}\sqrt{81{a}^{4}{b}^{4}⋅2a} & \ \text{Factor perfect square from radicand}. \\ \sqrt{81{a}^{4}{b}^{4}}⋅\sqrt{2a} & \ \text{Write radical expression as product of radical expressions}. \\ 9{a}^{2}{b}^{2}\sqrt{2a} & \ \text{Simplify}.\end{array}\)
Example

Try it.

Simplify the radical expression.
\(\sqrt{12}⋅\sqrt{3}\)

Solution

\(\begin{array}{ll}\sqrt{12⋅3} & \ \text{Express the product as a single radical expression}. \\ \sqrt{36} & \ \text{Simplify}. \\ 6 & \end{array}\)

Using the Quotient Rule to Simplify Square Roots

Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. We can rewrite \(\sqrt{\frac{5}{2}}\) as \(\frac{\sqrt{5}}{\sqrt{2}}.\)

Example

Try it.

Simplify the radical expression.

\(\sqrt{\frac{5}{36}}\)

Solution

\(\begin{array}{ll}\frac{\sqrt{5}}{\sqrt{36}} & \ \text{Write as quotient of two radical expressions}. \\ \frac{\sqrt{5}}{6} & \ \text{Simplify denominator}.\end{array}\)

Example

Try it.

Simplify the radical expression.

\(\frac{\sqrt{234{x}^{11}y}}{\sqrt{26{x}^{7}y}}\)

Solution

\(\begin{array}{ll}\sqrt{\frac{234{x}^{11}y}{26{x}^{7}y}} & \ \text{Combine numerator and denominator into one radical expression}. \\ \sqrt{9{x}^{4}} & \ \text{Simplify fraction}. \\ 3{x}^{2}\ & \ \text{Simplify square root}.\end{array}\)

Adding and Subtracting Square Roots

We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. For example, the sum of \(\sqrt{2}\) and \(3\sqrt{2}\) is \(4\sqrt{2}.\) However, it is often possible to simplify radical expressions, and that may change the radicand. The radical expression \(\sqrt{18}\) can be written with a \(2\) in the radicand, as \(3\sqrt{2},\) so \(\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}.\)

Example

Try it.

Add \(5\sqrt{12}+2\sqrt{3}.\)

Solution

We can rewrite \(5\sqrt{12}\) as \(5\sqrt{4\cdot 3}.\) According the product rule, this becomes \(5\sqrt{4}\sqrt{3}.\) The square root of \(\sqrt{4}\) is 2, so the expression becomes \(5(2)\sqrt{3},\) which is \(10\sqrt{3}.\) Now the terms have the same radicand so we can add.

\(10\sqrt{3}+2\sqrt{3}=12\sqrt{3}\)

Example

Try it.

Subtract \(20\sqrt{72{a}^{3}{b}^{4}c}\ -14\sqrt{8{a}^{3}{b}^{4}c}.\)

Solution

Factor 9 out of the first term so that both terms have equal radicands.

\[\begin{array}{lllll}20\sqrt{72{a}^{3}{b}^{4}c} & = & 20\sqrt{9⋅8{a}^{3}{b}^{4}c} & = & 20\sqrt{9}\sqrt{8{a}^{3}{b}^{4}c} \\ & = & 20(3)\sqrt{8{a}^{3}{b}^{4}c} & = & 60\sqrt{8{a}^{3}{b}^{4}c}\end{array}\]

So

\[\begin{array}{l}20\sqrt{72{a}^{3}{b}^{4}c}-14\sqrt{8{a}^{3}{b}^{4}c} \\ =60\sqrt{8{a}^{3}{b}^{4}c}-14\sqrt{8{a}^{3}{b}^{4}c} \\ =46\sqrt{8{a}^{3}{b}^{4}c}\end{array}\]

Rationalizing Denominators

When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator.

We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical.

For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is \(b\sqrt{c},\) multiply by \(\frac{\sqrt{c}}{\sqrt{c}}.\)

For a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator. If the denominator is \(a+b\sqrt{c},\) then the conjugate is \(a-b\sqrt{c}.\)

Example

Try it.

Write \(\frac{2\sqrt{3}}{3\sqrt{10}}\) in simplest form.

Solution

The radical in the denominator is \(\sqrt{10}.\) So multiply the fraction by \(\frac{\sqrt{10}}{\sqrt{10}}.\) Then simplify.

\[\begin{array}{l}\frac{2\sqrt{3}}{3\sqrt{10}}⋅\frac{\sqrt{10}}{\sqrt{10}}\ \\ \frac{2\sqrt{30}}{30}\ \\ \frac{\sqrt{30}}{15}\end{array}\]
Example

Try it.

Write \(\frac{4}{1+\sqrt{5}}\) in simplest form.

Solution

Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of \(1+\sqrt{5}\) is \(1-\sqrt{5}.\) Then multiply the fraction by \(\frac{1-\sqrt{5}}{1-\sqrt{5}}.\)

\[\begin{array}{ll}\frac{4}{1+\sqrt{5}}⋅\frac{1-\sqrt{5}}{1-\sqrt{5}} & \\ \frac{4-4\sqrt{5}}{-4} & \ \text{Use the distributive property}. \\ \sqrt{5}-1 & \ \text{Simplify}.\end{array}\]

Using Rational Roots

Although square roots are the most common rational roots, we can also find cube roots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number.

Suppose we know that \({a}^{3}=8.\) We want to find what number raised to the 3rd power is equal to 8. Since \({2}^{3}=8,\) we say that 2 is the cube root of 8.

The nth root of \(a\) is a number that, when raised to the nth power, gives \(a.\) For example, \(-3\) is the 5th root of \(-243\) because \({(-3)}^{5}=-243.\) If \(a\) is a real number with at least one nth root, then the principal nth root of \(a\) is the number with the same sign as \(a\) that, when raised to the nth power, equals \(a.\)

The principal nth root of \(a\) is written as \(\sqrt[n]{a},\) where \(n\) is a positive integer greater than or equal to 2. In the radical expression, \(n\) is called the index of the radical.

Example

Try it.

Simplify each of the following:

  1. ⓐ \(\sqrt[5]{-32}\)
  2. ⓑ \(\sqrt[4]{4}⋅\sqrt[4]{1,024}\)
  3. ⓒ \(-\sqrt[3]{\frac{8{x}^{6}}{125}}\)
  4. ⓓ \(8\sqrt[4]{3}-\sqrt[4]{48}\)
Solution
  1. ⓐ \(\sqrt[5]{-32}=-2\\) because \(\ {(-2)}^{5}=-32\)
  2. ⓑFirst, express the product as a single radical expression. \(\ \sqrt[4]{4,096}=8\\) because \(\ {8}^{4}=4,096\)
  3. ⓒ \(\begin{array}{ll}\frac{-\sqrt[3]{8{x}^{6}}}{\sqrt[3]{125}} & \ \text{Write as quotient of two radical expressions}. \\ \frac{-2{x}^{2}}{5} & \ \text{Simplify}.\end{array}\)
  4. ⓓ \(\begin{array}{ll}8\sqrt[4]{3}-2\sqrt[4]{3} & \ \text{Simplify to get equal radicands}. \\ 6\sqrt[4]{3} & \ \text{Add}.\end{array}\)

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

  • The principal square root of a number \(a\) is the nonnegative number that when multiplied by itself equals \(a.\) See .
  • If \(a\) and \(b\) are nonnegative, the square root of the product \(ab\) is equal to the product of the square roots of \(a\) and \(b\) See and .
  • If \(a\) and \(b\) are nonnegative, the square root of the quotient \(\frac{a}{b}\) is equal to the quotient of the square roots of \(a\) and \(b\) See and .
  • We can add and subtract radical expressions if they have the same radicand and the same index. See and .
  • Radical expressions written in simplest form do not contain a radical in the denominator. To eliminate the square root radical from the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. See and .
  • The principal nth root of \(a\) is the number with the same sign as \(a\) that when raised to the nth power equals \(a.\) These roots have the same properties as square roots. See .
  • Radicals can be rewritten as rational exponents and rational exponents can be rewritten as radicals. See and .
  • The properties of exponents apply to rational exponents. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. List each of the possible subsets of the 4 topics listed above using roster notation. Remember a subset is a collection of topics in which each topic listed is an element of the set Q we defined above. By including a topic, we are indicating that the student has mastered the topic.

  2. Verify in your work above you have listed all 16 subsets to set Q. Remember that a subset may contain all of the topics listed in Q.

  3. What formula could you use to help you determine the number of possible subsets? Remember that each topic could be mastered or not by a student. Show below that your formula would be equal to 16 for a list of 4 topics.

  4. Now use the formula you found in #3 to find the number of subsets possible if we include all 32 student-learning outcomes.

    Hint: In evaluating exponential terms, the function values increase very rapidly. To display very large (or very small) values, a calculator will use scientific notation. For example: 2.56 E6 is telling you to move the decimal point 6 places to the right and to insert zeros where you have missing values.

    For example: 2.56 E6 = 2,560,000 or 2 million, five hundred, sixty thousand.

  5. The subsets you created in #1 are referred to as knowledge spaces in the field of learning science. In this context mastery of one concept may depend on your mastery of another.

    List one skill in mathematics that would help to master each of the following SLO’s:

    • A = Graph the basic functions listed in the library of functions.
    • B = Find the domain of a function defined by an equation.
    • C = Create a new function through composition of functions.
    • D = Find linear functions that model data sets.

  6. Mastery of what are called linchpin topics will make it easier to learn other topics. For example, the ability to solve linear equations with variables on both sides can “unlock” a whole set of new skills for a student to master.

    List 3 other linchpin topics that would help you to master this math course. Discuss these with others in your class. Did they identify the same topics?

  7. A corequisite course in mathematics is designed to provide support to a student by reviewing linchpin topics right when and where students need the help. Review of these important foundational ideas allow the learner to move on and master the student learning objectives for the course.

    Brainstorm ideas with your classmates about ways this corequisite support course could help you in your learning.

  8. Evaluate each expression.

    1. ⓐ \(\sqrt{100}\)
    2. ⓑ \(\sqrt{\sqrt{16}}\)
    3. ⓒ \(\sqrt{25+144}\)
    4. ⓓ \(\sqrt{49}-\sqrt{81}\)
    Rivela la risposta
    1. ⓐ \(\sqrt{100}=10\) because \({10}^{2}=100\)
    2. ⓑ \(\sqrt{\sqrt{16}}=\sqrt{4}=2\) because \({4}^{2}=16\) and \({2}^{2}=4\)
    3. ⓒ \(\sqrt{25+144}=\sqrt{169}=13\) because \({13}^{2}=169\)
    4. ⓓ \(\sqrt{49}-\sqrt{81}=7-9=-2\) because \({7}^{2}=49\) and \({9}^{2}=81\)

  9. Evaluate each expression.

    1. ⓐ \(\sqrt{225}\)
    2. ⓑ \(\sqrt{\sqrt{81}}\)
    3. ⓒ \(\sqrt{25-9}\)
    4. ⓓ \(\sqrt{36}+\sqrt{121}\)
    Rivela la risposta
    1. ⓐ \(15\)
    2. ⓑ \(3\)
    3. ⓒ \(4\)
    4. ⓓ \(17\)

  10. Simplify the radical expression.

    1. ⓐ \(\sqrt{300}\)
    2. ⓑ \(\sqrt{162{a}^{5}{b}^{4}}\)
    Rivela la risposta
    1. ⓐ
      \(\begin{array}{ll}\sqrt{100⋅3} & \ \text{Factor perfect square from radicand}. \\ \sqrt{100}⋅\sqrt{3} & \ \text{Write radical expression as product of radical expressions}. \\ 10\sqrt{3} & \ \text{Simplify}.\end{array}\)
    2. ⓑ
      \(\begin{array}{ll}\sqrt{81{a}^{4}{b}^{4}⋅2a} & \ \text{Factor perfect square from radicand}. \\ \sqrt{81{a}^{4}{b}^{4}}⋅\sqrt{2a} & \ \text{Write radical expression as product of radical expressions}. \\ 9{a}^{2}{b}^{2}\sqrt{2a} & \ \text{Simplify}.\end{array}\)

  11. Simplify \(\sqrt{50{x}^{2}{y}^{3}z}.\)

    Rivela la risposta

    \(5|x||y|\sqrt{2yz}.\) Notice the absolute value signs around x and y? That’s because their value must be positive!

  12. Simplify the radical expression.
    \(\sqrt{12}⋅\sqrt{3}\)

    Rivela la risposta

    \(\begin{array}{ll}\sqrt{12⋅3} & \ \text{Express the product as a single radical expression}. \\ \sqrt{36} & \ \text{Simplify}. \\ 6 & \end{array}\)

  13. Simplify \(\sqrt{50x}⋅\sqrt{2x}\) assuming \(x>0.\)

    Rivela la risposta

    \(10|x|\)

  14. Simplify the radical expression.

    \(\sqrt{\frac{5}{36}}\)

    Rivela la risposta

    \(\begin{array}{ll}\frac{\sqrt{5}}{\sqrt{36}} & \ \text{Write as quotient of two radical expressions}. \\ \frac{\sqrt{5}}{6} & \ \text{Simplify denominator}.\end{array}\)

  15. Simplify \(\sqrt{\frac{2{x}^{2}}{9{y}^{4}}}.\)

    Rivela la risposta

    \(\frac{x\sqrt{2}}{3{y}^{2}}.\) We do not need the absolute value signs for \({y}^{2}\) because that term will always be nonnegative.

  16. Simplify the radical expression.

    \(\frac{\sqrt{234{x}^{11}y}}{\sqrt{26{x}^{7}y}}\)

    Rivela la risposta

    \(\begin{array}{ll}\sqrt{\frac{234{x}^{11}y}{26{x}^{7}y}} & \ \text{Combine numerator and denominator into one radical expression}. \\ \sqrt{9{x}^{4}} & \ \text{Simplify fraction}. \\ 3{x}^{2}\ & \ \text{Simplify square root}.\end{array}\)

  17. Simplify \(\frac{\sqrt{9{a}^{5}{b}^{14}}}{\sqrt{3{a}^{4}{b}^{5}}}.\)

    Rivela la risposta

    \({b}^{4}\sqrt{3ab}\)

  18. Add \(5\sqrt{12}+2\sqrt{3}.\)

    Rivela la risposta

    We can rewrite \(5\sqrt{12}\) as \(5\sqrt{4\cdot 3}.\) According the product rule, this becomes \(5\sqrt{4}\sqrt{3}.\) The square root of \(\sqrt{4}\) is 2, so the expression becomes \(5(2)\sqrt{3},\) which is \(10\sqrt{3}.\) Now the terms have the same radicand so we can add.

    \(10\sqrt{3}+2\sqrt{3}=12\sqrt{3}\)

  19. Add \(\sqrt{5}+6\sqrt{20}.\)

    Rivela la risposta

    \(13\sqrt{5}\)

  20. Subtract \(20\sqrt{72{a}^{3}{b}^{4}c}\ -14\sqrt{8{a}^{3}{b}^{4}c}.\)

    Rivela la risposta

    Factor 9 out of the first term so that both terms have equal radicands.

    \[\begin{array}{lllll}20\sqrt{72{a}^{3}{b}^{4}c} & = & 20\sqrt{9⋅8{a}^{3}{b}^{4}c} & = & 20\sqrt{9}\sqrt{8{a}^{3}{b}^{4}c} \\ & = & 20(3)\sqrt{8{a}^{3}{b}^{4}c} & = & 60\sqrt{8{a}^{3}{b}^{4}c}\end{array}\]

    So

    \[\begin{array}{l}20\sqrt{72{a}^{3}{b}^{4}c}-14\sqrt{8{a}^{3}{b}^{4}c} \\ =60\sqrt{8{a}^{3}{b}^{4}c}-14\sqrt{8{a}^{3}{b}^{4}c} \\ =46\sqrt{8{a}^{3}{b}^{4}c}\end{array}\]

  21. Subtract \(3\sqrt{80x}\ -4\sqrt{45x}.\)

    Rivela la risposta

    \(0\)

  22. Write \(\frac{2\sqrt{3}}{3\sqrt{10}}\) in simplest form.

    Rivela la risposta

    The radical in the denominator is \(\sqrt{10}.\) So multiply the fraction by \(\frac{\sqrt{10}}{\sqrt{10}}.\) Then simplify.

    \[\begin{array}{l}\frac{2\sqrt{3}}{3\sqrt{10}}⋅\frac{\sqrt{10}}{\sqrt{10}}\ \\ \frac{2\sqrt{30}}{30}\ \\ \frac{\sqrt{30}}{15}\end{array}\]

  23. Write \(\frac{12\sqrt{3}}{\sqrt{2}}\) in simplest form.

    Rivela la risposta

    \(6\sqrt{6}\)

  24. Write \(\frac{4}{1+\sqrt{5}}\) in simplest form.

    Rivela la risposta

    Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of \(1+\sqrt{5}\) is \(1-\sqrt{5}.\) Then multiply the fraction by \(\frac{1-\sqrt{5}}{1-\sqrt{5}}.\)

    \[\begin{array}{ll}\frac{4}{1+\sqrt{5}}⋅\frac{1-\sqrt{5}}{1-\sqrt{5}} & \\ \frac{4-4\sqrt{5}}{-4} & \ \text{Use the distributive property}. \\ \sqrt{5}-1 & \ \text{Simplify}.\end{array}\]

  25. Write \(\frac{7}{2+\sqrt{3}}\) in simplest form.

    Rivela la risposta

    \(14-7\sqrt{3}\)

  26. Simplify each of the following:

    1. ⓐ \(\sqrt[5]{-32}\)
    2. ⓑ \(\sqrt[4]{4}⋅\sqrt[4]{1,024}\)
    3. ⓒ \(-\sqrt[3]{\frac{8{x}^{6}}{125}}\)
    4. ⓓ \(8\sqrt[4]{3}-\sqrt[4]{48}\)
    Rivela la risposta
    1. ⓐ \(\sqrt[5]{-32}=-2\\) because \(\ {(-2)}^{5}=-32\)
    2. ⓑFirst, express the product as a single radical expression. \(\ \sqrt[4]{4,096}=8\\) because \(\ {8}^{4}=4,096\)
    3. ⓒ \(\begin{array}{ll}\frac{-\sqrt[3]{8{x}^{6}}}{\sqrt[3]{125}} & \ \text{Write as quotient of two radical expressions}. \\ \frac{-2{x}^{2}}{5} & \ \text{Simplify}.\end{array}\)
    4. ⓓ \(\begin{array}{ll}8\sqrt[4]{3}-2\sqrt[4]{3} & \ \text{Simplify to get equal radicands}. \\ 6\sqrt[4]{3} & \ \text{Add}.\end{array}\)

  27. Simplify.

    1. ⓐ \(\sqrt[3]{-216}\)
    2. ⓑ \(\frac{3\sqrt[4]{80}}{\sqrt[4]{5}}\)
    3. ⓒ \(6\sqrt[3]{9,000}+7\sqrt[3]{576}\)
    Rivela la risposta
    1. ⓐ \(-6\)
    2. ⓑ \(6\)
    3. ⓒ \(88\sqrt[3]{9}\)

  28. Write \({343}^{\frac{2}{3}}\) as a radical. Simplify.

    Rivela la risposta

    The 2 tells us the power and the 3 tells us the root.

    \({343}^{\frac{2}{3}}={(\sqrt[3]{343})}^{2}=\sqrt[3]{{343}^{2}}\)

    We know that \(\sqrt[3]{343}=7\) because \({7}^{3}=343.\) Because the cube root is easy to find, it is easiest to find the cube root before squaring for this problem. In general, it is easier to find the root first and then raise it to a power.

    \[{343}^{\frac{2}{3}}={(\sqrt[3]{343})}^{2}={7}^{2}=49\]

  29. Write \({9}^{\frac{5}{2}}\) as a radical. Simplify.

    Rivela la risposta

    \({(\sqrt{9})}^{5}={3}^{5}=243\)

  30. Write \(\frac{4}{\sqrt[7]{{a}^{2}}}\) using a rational exponent.

    Rivela la risposta

    The power is 2 and the root is 7, so the rational exponent will be \(\frac{2}{7}.\) We get \(\frac{4}{{a}^{\frac{2}{7}}}.\) Using properties of exponents, we get \(\frac{4}{\sqrt[7]{{a}^{2}}}=4{a}^{\frac{-2}{7}}.\)

  31. Write \(x\sqrt{{(5y)}^{9}}\) using a rational exponent.

    Rivela la risposta

    \(x{(5y)}^{\frac{9}{2}}\)

  32. Simplify:

    1. ⓐ \(5(2{x}^{\frac{3}{4}})(3{x}^{\frac{1}{5}})\)
    2. ⓑ \({(\frac{16}{9})}^{-\frac{1}{2}}\)
    Rivela la risposta

    ⓐ
    \(\begin{array}{ll}30{x}^{\frac{3}{4}}{x}^{\frac{1}{5}} & \ \text{Multiply the coefficients}. \\ 30{x}^{\frac{3}{4}+\frac{1}{5}} & \ \text{Use properties of exponents}. \\ 30{x}^{\frac{19}{20}} & \ \text{Simplify}.\end{array}\)

    ⓑ
    \(\begin{array}{ll}{(\frac{9}{16})}^{\frac{1}{2}} & \ \text{Use definition of negative exponents}. \\ \sqrt{\frac{9}{16}} & \ \text{Rewrite as a radical}. \\ \frac{\sqrt{9}}{\sqrt{16}} & \ \text{Use the quotient rule}. \\ \frac{3}{4} & \ \text{Simplify}.\end{array}\)

  33. Simplify \({(8x)}^{\frac{1}{3}}(14{x}^{\frac{6}{5}}).\)

    Rivela la risposta

    \(28{x}^{\frac{23}{15}}\)

  34. What does it mean when a radical does not have an index? Is the expression equal to the radicand? Explain.

    Rivela la risposta

    When there is no index, it is assumed to be 2 or the square root. The expression would only be equal to the radicand if the index were 1.

  35. Where would radicals come in the order of operations? Explain why.

  36. Every number will have two square roots. What is the principal square root?

    Rivela la risposta

    The principal square root is the nonnegative root of the number.

  37. Can a radical with a negative radicand have a real square root? Why or why not?

  38. \(\sqrt{256}\)

    Rivela la risposta

    16

  39. \(\sqrt{\sqrt{256}}\)

  40. \(\sqrt{4(9+16)}\)

    Rivela la risposta

    10

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\neq
not equal
The two sides are different.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i
imaginary unit
i² = −1.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Radicals and Rational Exponents

  1. Evaluate square roots.
  2. Use the product rule to simplify square roots.
  3. Use the quotient rule to simplify square roots.
  4. Add and subtract square roots.
  5. Rationalize denominators.
  6. Use rational roots.
  7. Investigate the discipline called learning science and the idea of a knowledge space.
  8. A = Graph the basic functions listed in the library of functions.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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