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Section A.10 Working with Radicals

In some situations, radical notation is more convenient to use than exponents. In these cases, we usually simplify radical expressions algebraically as much as possible before using a calculator to obtain decimal approximations.

Subsection Properties of Radicals

\(\sqrt[n]{a}=a^{1/n}\text{,}\) we can use the laws of exponents to derive two important properties that are useful in simplifying radicals.

Properties of Radicals.

  1. \(\displaystyle{\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}\text{, }\hphantom{blank000}\text{for } a, b \ge 0}\)

  2. \(\displaystyle{\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\text{, }\hphantom{blankblank}\text{for } a\ge 0,~~ b \gt 0}\)

As examples, you can verify that

\begin{equation*} \sqrt{36}=\sqrt{4}\sqrt{9} ~~~\text{ and }~~~ \sqrt[3]{\frac{1}{8}}=\frac{\sqrt[3]{1}}{\sqrt[3]{8}} \end{equation*}

Which of the following are true?

  1. Is \(\sqrt{36+64}= \sqrt{36}+\sqrt{64}\text{?}\)

  2. Is \(\sqrt[3]{8(64)}= \sqrt[3]{8}\sqrt[3]{64}\text{?}\)

  3. Is \(\sqrt{x^2+4}= x+2\text{?}\)

  4. Is \(\sqrt[3]{8x^3}= 2x\text{?}\)

Solution

The statements in (b) and (d) are true, and both are examples of the first property of radicals.

Statements (a) and (c) are false. In general, \(\sqrt[n]{a+b}\) is not equal to \(\sqrt[n]{a}+\sqrt[n]{b}\text{,}\) and \(\sqrt[n]{a-b}\) is not equal to \(\sqrt[n]{a}-\sqrt[n]{b}\text{.}\)

Subsection Simplifying Radicals

We use Property (1) to simplify radical expressions by factoring the radicand. For example, to simplify \(\sqrt[3]{108}\text{,}\) we look for perfect cubes that divide evenly into \(108\text{.}\) The easiest way to do this is to try the perfect cubes in order:

\begin{equation*} 1, ~8, ~27, ~64, ~125, \ldots \end{equation*}

and so on, until we find one that is a factor. For this example, we find that \(108 = 27 \cdot 4\text{.}\) Using Property (1), we write

\begin{equation*} \sqrt[3]{108}=\sqrt[3]{27}\sqrt[3]{4} \end{equation*}

Simplify the first factor to find

\begin{equation*} \sqrt[3]{108}= 3 \sqrt[3]{4} \end{equation*}

This expression is considered simpler than the original radical because the new radicand, \(4\text{,}\) is smaller than the original, \(108\text{.}\)

We can also simplify radicals containing variables. If the exponent on the variable is a multiple of the index, we can extract the variable from the radical. For instance,

\begin{equation*} \sqrt[3]{12}=x^{12/3}=x^4 \end{equation*}

(You can verify this by noting that \((x^4)^3 = x^{12}\text{.}\)) If the exponent on the variable is not a multiple of the index, we factor out the highest power that is a multiple. For example,

\begin{align*} \sqrt[3]{x^{11}} \amp = \sqrt[3]{x^9 \cdot x^2}\amp\amp\blert{\text{Apply Property (1).}}\\ \amp = \sqrt[3]{x^9}\cdot\sqrt[3]{2} \amp\amp\blert{\text{Simplify }\sqrt[3] {x^9}=x^{9/3}.} \\ \amp = x^3 \sqrt[3]{x^2} \end{align*}

Simplify each radical.

  1. \(\sqrt{18x^5}\)

  2. \(\sqrt[3]{24x^6y^8} \)

Solution

  1. The index of the radical is \(2\text{,}\) so we look for perfect square factors of \(18x^5\text{.}\) The factor \(9\) is a perfect square, and \(x^4\) has an exponent divisible by \(2\text{.}\) Thus,

    \begin{align*} \sqrt{18x^5} \amp = \sqrt{9x^4\cdot 2x}\amp\amp\blert{\text{Apply Property (1).}}\\ \amp = \sqrt{9x^4} \sqrt{2x}\amp\amp\blert{\text{Take square roots.}}\\ \amp = 3x^2\sqrt{2x} \end{align*}

  2. The index of the radical is \(3\text{,}\) so we look for perfect cube factors of \(24x^6 y^8\text{.}\) The factor \(8\) is a perfect cube, and \(x^6\) and \(y^6\) have exponents divisible by \(3\text{.}\) Thus,

    \begin{align*} \sqrt[3]{24x^6y^8} \amp = \sqrt[3]{8x^6 y^6 \cdot 3 y^2}\amp\amp\blert{\text{Apply Property (1).}}\\ \amp = \sqrt[3]{8x^6 y^6} \sqrt[3]{3 y^2}\amp\amp\blert{\text{Take cube roots.}}\\ \amp = 2x^2 y^2\sqrt[3]{3y^2} \end{align*}
Caution A.84.

Property (1) applies only to products under the radical, not to sums or differences. Thus, for example,

\begin{equation*} \sqrt{4\cdot 9}=\sqrt{4}\sqrt{9}=2\cdot 3, ~~~\text{ but }~~~\sqrt{4+ 9}\ne \sqrt{4}+\sqrt{9} \end{equation*}

and

\begin{equation*} \sqrt[3]{x^3y^6}=\sqrt[3]{x^3}\sqrt[3]{y^6}=xy^2, ~~~\text{ but }~~~\sqrt[3]{x^3-y^6}\ne \sqrt[3]{x^3}-\sqrt[3]{y^6} \end{equation*}

To simplify roots of fractions, we use Property (2), which allows us to write the expression as a quotient of two radicals.

  1. \(\displaystyle{\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}}\)

  2. \(\displaystyle{\sqrt[3]{\frac{5}{8}} =\frac{\sqrt[3]{5}}{\sqrt[3]{8}}=\frac{\sqrt[3]{5}}{2} }\)

We can also use Properties (1) and (2) to simplify products and quotients of radicals.

Simplify.

  1. \(\sqrt[4]{6x^2} \sqrt[4]{8x^3}\)

  2. \(\displaystyle{\frac{\sqrt[3]{16y^5}}{\sqrt[3]{y^2}}}\)

Solution

  1. First apply Property (1) to write the product as a single radical, then simplify.

    \begin{align*} \sqrt[4]{6x^2} \sqrt[4]{8x^3}\amp =\sqrt[4]{48x^5}\amp\amp\blert{\text{Factor out perfect fourth powers.}} \\ \amp = \sqrt[4]{16x^4}\sqrt[4]{3x}\amp\amp\blert{\text{Simplify.}}\\ \amp = 2x \sqrt[4]{3x} \end{align*}

  2. Apply Property (2) to write the quotient as a single radical.

    \begin{align*} \frac{\sqrt[3]{16y^5}}{\sqrt[3]{y^2}}\amp =\sqrt[3]{\frac{16y^5}{y^2}}\amp\amp\blert{\text{Reduce.}} \\ \amp = \sqrt[3]{16y^3}\amp\amp\blert{\text{Simplify: factor out perfect cubes.}}\\ \amp = \sqrt[3]{8y^3}\sqrt[3]{2}\\ \amp = 2y \sqrt[3]{2} \end{align*}

Subsection Sums and Differences of Radicals

You know that sums or differences of like terms can be combined by adding or subtracting their coefficients:

\begin{equation*} 3xy + 5xy = (3 + 5)xy = 8xy \end{equation*}

Like radicals, that is, radicals of the same index and radicand, can be combined in the same way.

  1. \(\begin{aligned}[t] 3\sqrt{3}+4\sqrt{3} \amp = (3+4)\sqrt{3}\\ \amp = 7\sqrt{3} \end{aligned}\)

  2. \(\begin{aligned}[t] 4\sqrt[3]{y}-6\sqrt[3]{y} \amp = (4-6)\sqrt[3]{y}\\ \amp = -2\sqrt[3]{y} \end{aligned}\)

Caution A.88.

  1. In Example A.87a, \(3\sqrt{3}+4\sqrt{3}\ne 7\sqrt{6} \text{.}\) Only the coefficients are added; the radicand does not change.

  2. Sums of radicals with different radicands or different indices cannot be combined. Thus,

    \begin{equation*} \sqrt{11}+\sqrt{5}\ne\sqrt{16} \end{equation*}
    \begin{equation*} \sqrt[3]{10x}-\sqrt[3]{2x}\ne\sqrt[3]{8x} \end{equation*}

    and

    \begin{equation*} \sqrt[3]{7}+\sqrt{7}\ne\sqrt[5]{7} \end{equation*}

    None of the expressions above can be simplified.

Subsection Products of Radicals

According to Property (1), radicals of the same index can be multiplied together.

Product of Radicals.
\begin{equation*} \sqrt[n]{a} \sqrt[n]{b}=\sqrt[n]{ab} \hphantom{blankblank} (a, b \ge 0) \end{equation*}

Thus, for example,

\begin{equation*} \sqrt{2}\sqrt{18}\sqrt{36}=6 ~~~\text{ and }~~~\sqrt[3]{2x}\sqrt[3]{4x^2}=\sqrt[3]{8x^3}=2x \end{equation*}

For products involving binomials, we can apply the distributive law.

  1. \(\begin{aligned}[t] \sqrt{3} \left(\sqrt{2x}+\sqrt{6} \right) \amp = \sqrt{3\cdot 2x}+ \sqrt{3\cdot 6}\\ \amp = \sqrt{6x}+\sqrt{18}=\sqrt{6x}+3\sqrt{2} \end{aligned}\)

  2. \(\begin{aligned}[t] \left(\sqrt{x}-\sqrt{y} \right)\left(\sqrt{x}+\sqrt{y} \right) \amp = \sqrt{x^2}+\sqrt{xy}-\sqrt{xy}-\sqrt{y^2}\\ \amp = x-y \end{aligned}\)

Subsection Rationalizing the Denominator

It is easier to work with radicals if there are no roots in the denominators of fractions. We can use the fundamental principle of fractions to remove radicals from the denominator. This process is called rationalizing the denominator. For square roots, we multiply the numerator and denominator of the fraction by the radical in the denominator.

Rationalize the denominator of each fraction.

  1. \(\displaystyle{\sqrt{\frac{1}{3}}}\)

  2. \(\displaystyle{\frac{\sqrt{2}}{\sqrt{50x}} }\)

Solution

  1. Apply Property (2) to write the radical as a quotient.

    \(\begin{aligned}[t] \sqrt{\frac{1}{3}}\amp = \frac{\sqrt{1}}{\sqrt{3}} \\ \amp =\frac{1}{\sqrt{3}} \amp\amp\blert{\text{Multiply numerator and denominator by } \sqrt{3}.}\\ \amp = \frac{1\cdot \alert{\sqrt{3}}}{\sqrt{3}\cdot \alert{\sqrt{3}}}\\ \amp = \frac{\sqrt{3}}{3} \end{aligned}\)

  2. It is always best to simplify the denominator before rationalizing.

    \(\begin{aligned}[t] \frac{\sqrt{2}}{\sqrt{50x}}\amp = \frac{\sqrt{2}}{5\sqrt{2x}} \amp\amp\blert{\text{Multiply numerator and denominator by } \sqrt{2x}.}\\ \amp = \frac{\sqrt{2}\cdot \alert{\sqrt{2x}}}{5\sqrt{2x}\cdot \alert{\sqrt{2x}}} \amp\amp\blert{\text{Simplify.}} \\ \amp = \frac{\sqrt{4x}}{5(2x)}\\ \amp =\frac{2\sqrt{x}}{10x} = \frac{\sqrt{x}}{5x}\\ \end{aligned}\)

If the denominator of a fraction is a binomial in which one or both terms is a radical, we can use a special building factor to rationalize it. First, recall that

\begin{equation*} (p - q) (p + q) = p^2 - q^2 \end{equation*}

where the product consists of perfect squares only. Each of the two factors \(p - q\) and \(p + q\) is said to be the conjugate of the other.

Now consider a fraction of the form

\begin{equation*} \frac{a}{b+\sqrt{c}} \end{equation*}

If we multiply the numerator and denominator of this fraction by the conjugate of the denominator, we get

\begin{equation*} \frac{a \alert{(b-\sqrt{c})}}{(b+\sqrt{c})\alert{(b-\sqrt{c})}}=\frac{ab-a\sqrt{c}}{b^2-(\sqrt{c})^2} =\frac{ab-a\sqrt{c}}{b^2-c} \end{equation*}

The denominator of the fraction no longer contains any radicals—it has been rationalized.

Multiplying numerator and denominator by the conjugate of the denominator also works on fractions of the form

\begin{equation*} \frac{a}{\sqrt{b}+c} ~~~\text{ and }~~~\frac{a}{\sqrt{b}+\sqrt{c}} \end{equation*}

We leave the verification of these cases as exercises.

Rationalize the denominator: \(\displaystyle{\frac{x}{\sqrt{2}+\sqrt{x}}} \text{.}\)

Solution

Multiply numerator and denominator by the conjugate of the denominator, \(\sqrt{2}-\sqrt{x} \text{.}\)

\begin{equation*} \frac{x (\alert{\sqrt{2}-\sqrt{x}}) }{(\sqrt{2}+\sqrt{x})(\alert{\sqrt{2}-\sqrt{x}} )} =\frac{x(\sqrt{2}-\sqrt{x})}{2-x} \end{equation*}

Subsection Simplifying \(\sqrt[n]{x^n} \)

Raising to a power is the inverse operation for extracting roots; that is,

\begin{equation*} \left(\sqrt[n]{a} \right)^n =a \end{equation*}

as long as \(\sqrt[n]{a} \) is a real number. For example,

\begin{equation*} \left(\sqrt[4]{16} \right)^4= 2^4 = 16,~~~\text{ and }~~~\left(\sqrt[3]{-125} \right)^3= (-5)^3 = -125 \end{equation*}

Now consider the power and root operations in the opposite order; is it true that \(\sqrt[n]{a^n}=a \text{?}\) If the index \(n\) is an odd number, then the statement is always true. For example,

\begin{equation*} \sqrt[3]{2^3}=\sqrt[3]{8}=2 ~~~\text{ and }~~~\sqrt[3]{(-2)^3}=\sqrt[3]{-8}=-2 \end{equation*}

However, if \(n\) is even, we must be careful. Recall that the principal root \(\sqrt[n]{x} \) is always positive, so if \(a\) is a negative number, it cannot be true that \(\sqrt[n]{a^n}=a \text{.}\) For example, if \(a = -3\text{,}\) then

\begin{equation*} \sqrt{(-3)^2}=\sqrt{9}=3 \end{equation*}

Instead, we see that, for even roots, \(\sqrt[n]{a^n}=\abs{a} \text{.}\)

We summarize our results in below.

Roots of Powers.

  1. If \(n\) is odd, \(\hphantom{blankblank}\sqrt[n]{a^n}=a \)

  2. If \(n\) is even, \(\hphantom{blankblank}\sqrt[n]{a^n}=\abs{a} \)

    In particular, \(\hphantom{blankblank}\sqrt{a^2}=\abs{a} \)

  1. \(\sqrt{16x^2}= 4\abs{x} \)

  2. \(\sqrt{(x-1)^2}= \abs{x-1} \)

Subsection Extraneous Solutions to Radical Equations

It is important to check the solution to a radical equation, because it is possible to introduce false, or extraneous, solutions when we square both sides of the equation. For example, the equation

\begin{equation*} \sqrt{x}=-5 \end{equation*}

has no solution, because \(\sqrt{x}\) is never a negative number. However, if we try to solve the equation by squaring both sides, we find

\begin{align*} (\sqrt{x})^\alert{2}\amp = (-5)^\alert{2}\\ x\amp = 25 \end{align*}

You can check that \(25\) is not a solution to the original equation, \(\sqrt{x}=-5 \text{,}\) because \(\sqrt{25} \) does not equal \(-5\text{.}\)

If each side of an equation is raised to an odd power, extraneous solutions will not be introduced. However, if we raise both sides to an even power, we should check each solution in the original equation.

Solve the equation \(\sqrt{x+2}+4=x \text{.}\)

Solution

First, isolate the radical expression on one side of the equation. (This will make it easier to square both sides.)

\begin{align*} \sqrt{x+2}\amp = x-4\amp\amp\blert{\text{Square both sides of the equation.}}\\ \left(\sqrt{x+2}\right)^\alert{2}\amp = (x-4)^\alert{2}\\ x+2\amp = x^2-8x+16 \amp\amp\blert{\text{Subtract }x + 2 \text{ from both sides.}}\\ x^2-9x+14\amp = 0\amp\amp\blert{\text{Factor the left side.}}\\ x=2\hphantom{000}\amp\text{or}\hphantom{000}x=7 \end{align*}

Check

Does \(\sqrt{\alert{2}+2}+4=\alert{2} \text{?}\)

\(\hphantom{blank}\blert{{\text{No; }2\text{ is not a solution.}}} \)

Does \(\sqrt{\alert{7}+2}+4=\alert{7} \text{?}\)

\(\hphantom{blank}\blert{{\text{Yes; }7\text{ is a solution.}}} \)

The apparent solution \(2\) is extraneous. The only solution to the original equation is \(7\text{.}\) We can verify the solution by graphing the equations

\begin{equation*} y_1=\sqrt{x+2} ~~~\text{ and }~~~ y_2=x-4 \end{equation*}

as shown at right. The graphs intersect in only one point, \((7, 3)\text{,}\) so there is only one solution, \(x=7\text{.}\)

graphs of both sides of radical equation
Caution A.94.

When we square both sides of an equation, it is not correct to square each term of the equation separately. Thus, in Example A.93, the original equation is not equivalent to

\begin{equation*} (\sqrt{x+2})^2+4^2=x^2 \end{equation*}

This is because \((a + b)^2 \ne a^2 + b^2\text{.}\) Instead, we must square the entire left side of the equation as a binomial, like this,

\begin{equation*} (\sqrt{x+2}+4)^2=x^2 \end{equation*}

or we may proceed as shown in Example A.93.

Subsection Equations with More than One Radical

Sometimes it is necessary to square both sides of an equation more than once in order to eliminate all the radicals.

Solve \(~\sqrt{x-7}+\sqrt{x}=7 \text{.}\)

Solution

First, isolate the more complicated radical on one side of the equation. (This will make it easier to square both sides.) We will subtract \(\sqrt{x} \) from both sides.

\begin{equation*} \sqrt{x-7}=7-\sqrt{x} \end{equation*}

Now square each side to remove one radical. Be careful when squaring the binomial \(7-\sqrt{x} \text{.}\)

\begin{align*} (\sqrt{x-7})^\alert{2}\amp =(7-\sqrt{x})^\alert{2}\\ x-7\amp = 49-14\sqrt{x}+x \end{align*}

Collect like terms, and isolate the radical on one side of the equation.

\begin{align*} -56 \amp = -14\sqrt{x}\amp\amp\blert{\text{Divide both sides by }-14.}\\ 4 \amp = \sqrt{x} \end{align*}

Now square again to obtain

\begin{align*} (4)^\alert{2} \amp = (\sqrt{x})^\alert{2}\\ 16 \amp = x \end{align*}

Check

Does \(\sqrt{\alert{16}-7}+\sqrt{\alert{16}}=7\text{?}\)

\(\hphantom{blank}\blert{\text{Yes. The solution is } 16.}\)

Caution A.96.

Recall that we cannot solve a radical equation by squaring each term separately. In other words, it is incorrect to begin Example 95 by writing

\begin{equation*} (\sqrt{x-7})^2+(\sqrt{x})^2=7^2 \end{equation*}

We must square the entire expression on each side of the equal sign as one piece.

Subsection Section Summary

Subsubsection Vocabulary

Look up the definitions of new terms in the Glossary.

  • Radical

  • Extraneous

  • Rationalize

  • Conjugate

  • Like radicals

  • Radicand

  • Index

Subsubsection SKILLS

Practice each skill in the exercises listed.

  1. Simplify radicals: #1–6

  2. Simplify products and quotients of radcials: #7–10

  3. Combine like radicals: #11–18

  4. Multiply radical expressions: #19–36

  5. Rationalize the denominator: #37–50

  6. Simplify \(\sqrt[n]{a^n} \text{:}\) #51–54

  7. Solve radical equations: #55–80

Exercises Exercises A.10

For Problems 1-6, simplify. Assume that all variables represent positive numbers.

1.

  1. \(\sqrt{18} \)

  2. \(\sqrt[3]{24} \)

  3. \(-\sqrt[4]{64} \)

2.

  1. \(\sqrt{50} \)

  2. \(\sqrt[3]{54} \)

  3. \(-\sqrt[4]{162} \)

3.

  1. \(\sqrt{60,000} \)

  2. \(\sqrt[3]{900,000} \)

  3. \(\sqrt[3]{\dfrac{-40}{27}} \)

4.

  1. \(\sqrt{800,000} \)

  2. \(\sqrt[3]{24,000} \)

  3. \(\sqrt[4]{\dfrac{80}{625}} \)

5.

  1. \(\sqrt[3]{x^{10}} \)

  2. \(\sqrt{27z^3} \)

  3. \(\sqrt[4]{48a^9} \)

6.

  1. \(\sqrt[3]{y^{16}} \)

  2. \(\sqrt{12t^5} \)

  3. \(\sqrt[3]{81b^8} \)

For Problems 7-10, simplify.

7.

  1. \(-\sqrt{18s}\sqrt{2s^3} \)

  2. \(\sqrt[3]{7h^2}\sqrt[3]{-49h} \)

  3. \(\sqrt{16-4x^2} \)

8.

  1. \(\sqrt{3w^3}\sqrt{27w^3} \)

  2. \(-\sqrt[4]{2m^3}\sqrt[4]{8m} \)

  3. \(\sqrt{9Y^2+18} \)

9.

  1. \(\sqrt[3]{8A^3+A^6} \)

  2. \(\dfrac{\sqrt{45x^3y^3}}{\sqrt{5y}} \)

  3. \(\dfrac{\sqrt[3]{8b^7}}{\sqrt[3]{a^6b^2}} \)

10.

  1. \(\sqrt[3]{b^9-27b^3} \)

  2. \(\dfrac{\sqrt{98x^2y^3}}{\sqrt{xy}} \)

  3. \(\dfrac{\sqrt[3]{16r^4}}{\sqrt[3]{4t^3}} \)

For Problems 11-18, simplify and combine like terms.

11.

\(3\sqrt{7}+2\sqrt{7}\)

12.

\(5\sqrt{2}-3\sqrt{2}\)

13.

\(4\sqrt{3}-\sqrt{27}\)

14.

\(\sqrt{75}+2\sqrt{3}\)

15.

\(\sqrt{50x}+\sqrt{32x}\)

16.

\(\sqrt{8y}-\sqrt{18y}\)

17.

\(3\sqrt[3]{16}-\sqrt[3]{2}-2\sqrt[3]{54}\)

18.

\(\sqrt[3]{81}+2\sqrt[3]{24}-3\sqrt[3]{3}\)

For Problems 19-32, multiply.

19.

\(2(3-\sqrt{5})\)

20.

\(5(2-\sqrt{7})\)

21.

\(\sqrt{2}(\sqrt{6}+\sqrt{10})\)

22.

\(\sqrt{3}(\sqrt{12}-\sqrt{15})\)

23.

\(\sqrt[3]{2}(\sqrt[3]{20}-2\sqrt[3]{12})\)

24.

\(\sqrt[3]{3}(2\sqrt[3]{18}+\sqrt[3]{36})\)

25.

\((\sqrt{x}-3)(\sqrt{x}+3)\)

26.

\((2+\sqrt{x})(2-\sqrt{x})\)

27.

\((\sqrt{2}-\sqrt{3})(\sqrt{2}+2\sqrt{3})\)

28.

\((\sqrt{3}-\sqrt{5})(2\sqrt{3}+\sqrt{5})\)

29.

\((\sqrt{5}-\sqrt{2})^2\)

30.

\((\sqrt{2}-2\sqrt{3})^2\)

31.

\((\sqrt{a}-2\sqrt{b})^2\)

32.

\((\sqrt{2a}-2\sqrt{b})(\sqrt{2a}+2\sqrt{b})\)

For Problems 33-36, verify by substitution that the number is a solution of the quadratic equation.

33.

\(x^2-2x-2=0 \text{, }~1+\sqrt{3}\)

34.

\(x^2+4x-1=0 \text{, }~-2+\sqrt{5}\)

35.

\(x^2+6x-9=0 \text{, }~-3+3\sqrt{2}\)

36.

\(4x^2-20x+22=0 \text{, }~\dfrac{5-\sqrt{3}}{2}\)

For Problems 37-50, rationalize the denominator.

37.

\(\dfrac{6}{\sqrt{3}}\)

38.

\(\dfrac{10}{\sqrt{5}}\)

39.

\(\sqrt{\dfrac{7x}{18}}\)

40.

\(\sqrt{\dfrac{27x}{20}}\)

41.

\(\sqrt{\dfrac{2a}{b}}\)

42.

\(\sqrt{\dfrac{5p}{q}}\)

43.

\(\dfrac{2\sqrt{3}}{\sqrt{2k}}\)

44.

\(\dfrac{6\sqrt{2}}{\sqrt{3v}}\)

45.

\(\dfrac{4}{1+\sqrt{3}}\)

46.

\(\dfrac{3}{7-\sqrt{2}}\)

47.

\(\dfrac{x}{x-\sqrt{3}}\)

48.

\(\dfrac{y}{\sqrt{5}-y}\)

49.

\(\dfrac{\sqrt{6}-3}{2-\sqrt{6}}\)

50.

\(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

51.

Use your calculator to graph each function, and explain the result.

  1. \(y=\sqrt{x^2} \)

  2. \(y=\sqrt[3]{x^3} \)

52.

Use your calculator to graph each function, and explain the result.

  1. \(y=(x^4)^{1/4} \)

  2. \(y=(x^5)^{1/5} \)

For Problems 53-54, do not assume that variables represent positive numbers. Use absolute value bars as necessary to simplify the radicals.

53.

  1. \(\sqrt{4x^2} \)

  2. \(\sqrt{(x-5)^2} \)

  3. \(\sqrt{x^2-6x+9} \)

54.

  1. \(\sqrt{9x^2y^4} \)

  2. \(\sqrt{(2x-1)^2} \)

  3. \(\sqrt{9x^2-6x+1} \)

For Problems 55-78, solve

55.

\(\sqrt{x}-5=3\)

56.

\(\sqrt{x}-4=1\)

57.

\(\sqrt{y+6}=2\)

58.

\(\sqrt{y-3}=5\)

59.

\(4\sqrt{z}-8=-2\)

60.

\(-3\sqrt{z}+14=8\)

61.

\(5+2\sqrt{6-2w}=13\)

62.

\(8-3\sqrt{9+2w}=-7\)

63.

\(3z+4=\sqrt{3z+10}\)

64.

\(2x-3=\sqrt{7x-3}\)

65.

\(2x+1=\sqrt{10x+5}\)

66.

\(4x+5=\sqrt{3x+4}\)

67.

\(\sqrt{y+4}=y-8\)

68.

\(4\sqrt{x-4}=x\)

69.

\(\sqrt{2y-1}=\sqrt{3y-6}\)

70.

\(\sqrt{4y+1}=\sqrt{6y-3}\)

71.

\(\sqrt{x-3}\sqrt{x}=2\)

72.

\(\sqrt{x}\sqrt{x-5}=6\)

73.

\(\sqrt{y+4}=\sqrt{y+20}-2\)

74.

\(4\sqrt{y}+\sqrt{1+16y}=5\)

75.

\(\sqrt{x}+\sqrt{2}=\sqrt{x+2}\)

76.

\(\sqrt{4x+17}=4-\sqrt{x+1}\)

77.

\(\sqrt{5+x}+\sqrt{x}=5\)

78.

\(\sqrt{y+7}+\sqrt{y+4}=3\)

79.

Explain why the following first step for solving the radical equation is incorrect:

\begin{align*} \sqrt{x-5}+\sqrt{2x-1}\amp=8 \\ (x-5)+(2x-1)\amp =64 \end{align*}
80.

Explain why the following first step for solving the radical equation is incorrect:

\begin{align*} \sqrt{x+2}+1 \amp=\sqrt{2x-3} \\ (x+2)+1\amp =2x-3 \end{align*}

For Problems 81-84, write the complex fraction as a simple fraction in lowest terms, and rationalize the denominator.

81.

\(\displaystyle{\frac{\dfrac{2}{\sqrt{7}}}{1-\dfrac{\sqrt{3}}{\sqrt{7}}}}\)

82.

\(\displaystyle{\frac{\dfrac{1}{4}}{\dfrac{\sqrt{5}}{2\sqrt{2}}-\dfrac{\sqrt{3}}{2\sqrt{2}}}}\)

83.

\(\displaystyle{\frac{\dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}}{1-\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{\sqrt{2}}}}\)

84.

\(\displaystyle{\frac{\dfrac{1}{\sqrt{3}}-\dfrac{\sqrt{5}}{3}}{1+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{5}}{3}}}\)