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Learning Objectives

• Simplify radical expressions using the product and quotient rule for radicals.
• Evaluate given square root and cube root functions.

An algebraic expression that contains radicals is called a radical expression. We use the product and quotient rules to simplify them.

Example $$\PageIndex{1}$$

Simplify:

$$\sqrt[3]{8 y^{3}}$$

Solution:

Use the fact that $$\sqrt[n]{a^{n}}=a$$ when n is odd.

\begin{aligned} \sqrt[3]{8 y^{3}} &=\sqrt[3]{2^{3} \cdot y^{3}} \qquad\quad\color{Cerulean}{Apply\:the\:product\:rule\:for\:radicals.}\\ &=\sqrt[3]{2^{3}} \cdot \sqrt[3]{y^{3}}\quad\:\:\:\color{Cerulean}{Simplify.} \\ &=2 y \end{aligned}

$$2y$$

Example $$\PageIndex{2}$$

Simplify:

$$\sqrt{9 x^{2}}$$

Solution:

The square root has index 2; use the fact that $$\sqrt[n]{a^{n}}=a$$ when n is even.

\begin{aligned} \sqrt{9 x^{2}} &=\sqrt{3^{2} x^{2}}\qquad\quad\color{Cerulean}{Apply\:the\:product\:rule\:for\:radicals.} \\ &=\sqrt{3^{2}} \cdot \sqrt{x^{2}}\quad\:\color{Cerulean}{Simplify.} \\ &=3|x| \end{aligned}

Since x is a variable, it may represent a negative number. Thus we need to ensure that the result is positive by including the absolute value operator.

$$3|x|$$

Note

Typically, at this point beginning algebra texts note that all variables are assumed to be positive. If this is the case, then x in the previous example is positive and the absolute value operator is not needed. The example can be simplified as follows:

$$\sqrt{9x^{2}}=\sqrt{3^{2}x^{2}}=\sqrt{3^{2}}\cdot\sqrt{x^{2}}=3x$$

In this section, we will assume that all variables are positive. This allows us to focus on calculating n th roots without the technicalities associated with the principal n th root problem. For this reason, we will use the following property for the rest of the section:

$$\sqrt[n]{a^{n}}=a$$, if $$a≥0$$ n th root

When simplifying radical expressions, look for factors with powers that match the index.

Example $$\PageIndex{3}$$

Simplify:

$$\sqrt{18x^{3}y^{4}}$$

Solution:

Begin by determining the square factors of $$18, x^{3}$$, and $$y^{4}$$.

\ \begin{aligned} 18 &=2 \cdot \color{Cerulean}{3^{2}} \\ x^{3} &=\color{Cerulean}{x^{2}}\color{black}{ \cdot} x \\ y^{4} &=\color{Cerulean}{\left(y^{2}\right)^{2}} \end{aligned} \ \qquad\color{Cerulean}{Square\:factors}

Make these substitutions and then apply the product rule for radicals and simplify.

\begin{aligned} \sqrt{18 x^{3} y^{4}} &=\sqrt{\color{Cerulean}{2}\color{black}{ \cdot} 3^{2} \cdot x^{2} \cdot \color{Cerulean}{x}\color{black}{ \cdot}\left(y^{2}\right)^{2}}\qquad\qquad\color{Cerulean}{Apply\:the\:product\:rule\:for\:radicals.} \\ &=\sqrt{3^{2}} \cdot \sqrt{x^{2}} \cdot \sqrt{\left(y^{2}\right)^{2}} \cdot \color{black}{\sqrt{\color{Cerulean}{2 x}}}\quad\color{Cerulean}{Simplify.} \\ &=3 \cdot x \cdot y^{2} \cdot \sqrt{2 x} \\ &=3 x y^{2} \sqrt{2 x} \end{aligned}

$$3 x y^{2} \sqrt{2 x}$$

Example $$\PageIndex{4}$$

Simplify:

$$\sqrt{4a^{5}b^{6}}$$

Solution:

Begin by determining the square factors of $$4, a^{5}$$, and $$b^{6}$$.

$$\begin{array}{l}{4=\color{Cerulean}{2^{2}}} \\ {a^{5}=a^{2} \cdot a^{2} \cdot a=\color{Cerulean}{\left(a^{2}\right)^{2}}\color{black}{ \cdot} a} \\ {b^{6}=b^{3} \cdot b^{3}=\color{Cerulean}{\left(b^{3}\right)^{2}}}\end{array} \qquad\color{Cerulean}{Square\:factors}$$

Make these substitutions and then apply the product rule for radicals and simplify.

\begin{aligned} \sqrt{\frac{4 a^{5}}{b^{6}}} &=\sqrt{\frac{2^{2}\left(a^{2}\right)^{2} \cdot a}{\left(b^{3}\right)^{2}}}\qquad\qquad\color{Cerulean}{Apply\:the\:product\:and\:quotient\:rule\:for\:radicals.} \\ &=\frac{\sqrt{2^{2}} \cdot \sqrt{\left(a^{2}\right)^{2}} \cdot \sqrt{a}}{\sqrt{\left(b^{3}\right)^{2}}}\quad\color{Cerulean}{Simplify.} \\ &=\frac{2 a^{2} \sqrt{a}}{b^{3}} \end{aligned}

$$\frac{2 a^{2} \sqrt{a}}{b^{3}}$$

Example $$\PageIndex{5}$$

Simplify:

$$\sqrt[3]{80x^{5}y^{7}}$$

Solution:

Begin by determining the cubic factors of $$80, x^{5}$$, and $$y^{7}$$.

$$\begin{array}{l}{80=2^{4} \cdot 5=\color{Cerulean}{2^{3}}\color{black}{ \cdot} 2 \cdot 5} \\ {x^{5}=\color{Cerulean}{x^{3}}\color{black}{ \cdot} x^{2}} \\ {y^{7}=y^{6} \cdot y=\color{Cerulean}{\left(y^{2}\right)^{3}}\color{black}{ \cdot} y}\end{array} \qquad\color{Cerulean}{Cubic\:factors}$$

Make these substitutions and then apply the product rule for radicals and simplify.

\begin{aligned} \sqrt[3]{80 x^{5} y^{7}} &=\sqrt[3]{\color{Cerulean}{2^{3}}\color{black}{ \cdot} 2 \cdot 5 \cdot \color{Cerulean}{x^{3}}\color{black}{ \cdot} x^{2} \cdot\color{Cerulean}{\left(y^{2}\right)^{3}}\color{black}{ \cdot} y} \qquad\qquad\qquad\color{Cerulean}{Apply\:the\:product\:rule\:for\:radicals.}\\ &=\color{black}{\sqrt[3]{\color{Cerulean}{2^{3}}}} \cdot \color{black}{\sqrt[3]{\color{Cerulean}{x^{3}}}} \cdot \color{black}{\sqrt[3]{\color{Cerulean}{\left(y^{2}\right)^{3}}}} \cdot \sqrt[3]{2 \cdot 5 \cdot x^{2} \cdot y} \quad\:\:\color{Cerulean}{Simplify.} \\ &=2 \cdot x \cdot y^{2} \cdot \sqrt[3]{10 x^{2} y} \\ &=2 x y^{2} \sqrt[3]{10 x^{2} y} \end{aligned}

$$2 x y^{2} \sqrt[3]{10 x^{2} y}$$

Example $$\PageIndex{6}$$

Simplify:

$$\sqrt[3]{9x^{6}y^{3}z^{9}}$$

Solution:

The coefficient $$9=3^{2}$$ and thus does not have any perfect cube factors. It will be left as the only remaining radicand because all of the other factors are cubes, as illustrated below:

\begin{aligned} x^{6} &=\left(x^{2}\right)^{3} \\ y^{3} &=(y)^{3} \\ z^{9} &=\left(z^{3}\right)^{3} \end{aligned}\qquad \color{Cerulean}{Cubic\:factors}

Replace the variables with these equivalents, apply the product and quotient rule for radicals, and then simplify.

\begin{aligned} \sqrt[3]{\frac{9 x^{6}}{y^{3} z^{9}}} &=\sqrt[3]{\frac{3^{2} \cdot\left(x^{2}\right)^{3}}{y^{3} \cdot\left(z^{3}\right)^{3}}} \\ &=\frac{\sqrt[3]{3^{2}} \cdot \sqrt[3]{\left(x^{2}\right)^{3}}}{\sqrt[3]{y^{3}} \cdot \sqrt[3]{\left(z^{3}\right)^{3}}} \\ &=\frac{\sqrt[3]{3^{2}} \cdot x^{2}}{y \cdot z^{3}} \\ &=\frac{\sqrt[3]{9} \cdot x^{2}}{y \cdot z^{3}} \end{aligned}

$$\frac{\sqrt[3]{9} \cdot x^{2}}{y \cdot z^{3}}$$

Example $$\PageIndex{7}$$

Simplify:

$$\sqrt[4]{81a^{4}b^{5}}$$

Solution:

Determine all factors that can be written as perfect powers of 4. Here it is important to see that $$b^{5}=b^{4}⋅b$$. Hence the factor $$b$$ will be left inside the radical.

\begin{aligned} \sqrt[4]{81 a^{4} b^{5}} &=\sqrt[4]{3^{4} \cdot a^{4} \cdot b^{4} \cdot b} \\ &=\sqrt[4]{3^{4}} \cdot \sqrt[4]{a^{4}} \cdot \sqrt[4]{b^{4}} \cdot \sqrt[4]{b} \\ &=3 \cdot a \cdot b \cdot \sqrt[4]{b} \end{aligned}

$$3ab\sqrt[4]{b}$$

Example $$\PageIndex{8}$$

Simplify:

$$\sqrt[5]{-32 x^{3} y^{6} z^{5}}$$

Solution:

Notice that the variable factor x cannot be written as a power of 5 and thus will be left inside the radical. In addition, for $$y^{6}=y^{5}⋅y$$; the factor y will be left inside the radical as well.

\begin{aligned} \sqrt[5]{-32 x^{3} y^{6} z^{5}} &=\sqrt[5]{(-2)^{5} \cdot\color{Cerulean}{ x^{3}}\color{black}{ \cdot} y^{5} \cdot \color{Cerulean}{y}\color{black}{ \cdot} z^{5}} \\ &=\sqrt[5]{(-2)^{5}} \cdot \sqrt[5]{y^{5}} \cdot \sqrt[5]{z^{5}} \cdot \color{black}{\sqrt[5]{\color{Cerulean}{x^{3} \cdot y}}} \\ &=-2 \cdot y \cdot z \cdot \sqrt[5]{x^{3} \cdot y} \end{aligned}

$$-2yz\sqrt[5]{x^{3}y}$$

Exercise $$\PageIndex{1}$$

Simplify:

$$\sqrt{192x^{6}y^{7}z^{12}}$$

(Assume all variables are positive.)

$$8x^{3}y^{3}z^{6}\sqrt{3y}$$

Note

To easily simplify an n th root, we can divide the powers by the index.

$$\sqrt{a^{6}}=a^{3}$$, which is    $$a^{6÷2}= a^{3}$$ $$\sqrt[3]{b^{6}}=b^{2}$$, which is     $$b^{6÷3}=b^{2}$$ $$\sqrt[6]{c^{6}}=c$$, which is  $$c^{6÷6}=c^{1}$$

If the index does not divide into the power evenly, then we can use the quotient and remainder to simplify. For example,

$$\sqrt{a^{5}}=a^{2}⋅\sqrt{a}$$,  which is $$a^{5÷2}=a^{2}_{r\:1}$$ $$\sqrt[3]{b^{5}}=b⋅\sqrt[3]{b^{2}}$$,  which is $$b^{5÷3}=b^{1}_{r\:2}$$ $$\sqrt[5]{c^{14}}=c^{2}⋅\sqrt[5]{c^{4}}$$,  which is     $$c^{14÷5}=c^{2}_{r\:4}$$

The quotient is the exponent of the factor outside of the radical, and the remainder is the exponent of the factor left inside the radical.

We next review the distance formula. Given two points $$(x_{1}, y_{1})$$ and $$(x_{2}, y_{2})$$,

The distance, d, between them is given by the following formula:

Distance formula:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Recall that this formula was derived from the Pythagorean theorem.

Example $$\PageIndex{9}$$

Calculate the distance between $$(−4, 7)$$ and $$(2, 1)$$.

Solution:

Use the distance formula with the following points.

$$\begin{array}{ll}{\left(x_{1}, y_{1}\right)} & {\left(x_{2}, y_{2}\right)} \\ {(\color{Cerulean}{-4}\color{black}{,}\color{OliveGreen}{7}\color{black}{)}} & {(\color{Cerulean}{2}\color{black}{,}\color{OliveGreen}{1}\color{black}{)}}\end{array}$$

It is a good practice to include the formula in its general form before substituting values for the variables; this improves readability and reduces the probability of making errors.

\begin{aligned} d &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ &=\sqrt{(\color{Cerulean}{2}\color{black}{-}(\color{Cerulean}{-4}\color{black}{)})^{2}+(\color{OliveGreen}{1}\color{black}{-}\color{OliveGreen}{7}\color{black}{)}^{2}} \\ &=\sqrt{(2+4)^{2}+(1-7)^{2}} \\ &=\sqrt{(6)^{2}+(-6)^{2}} \\ &=\sqrt{72} \\ &=\sqrt{36 \cdot 2} \\ &=6 \sqrt{2} \end{aligned}

$$6\sqrt{2}$$ units

Example $$\PageIndex{10}$$

The period, T, of a pendulum in seconds is given by the formula

$T=2 \pi \sqrt{\frac{L}{32}}$

where L represents the length of the pendulum in feet. If the length of a pendulum measures 6 feet, then calculate the period rounded off to the nearest tenth of a second.

Solution:

Substitute 6 for L and then simplify.

\begin{aligned} T &=2 \pi \sqrt{\frac{L}{32}} \\ &=2 \pi \sqrt{\frac{6}{32}}\quad\color{Cerulean}{Reduce.} \\ &=2 \pi \sqrt{\frac{3}{16}} \quad\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:radicals.}\\ &=2 \pi \frac{\sqrt{3}}{\sqrt{16}} \quad\color{Cerulean}{Simplify.}\\ &=\frac{2 \pi \sqrt{3}}{4}\quad\:\:\:\color{Cerulean}{Use\:a\:calculator.} \\ & \approx 2.7 \end{aligned}

The period is approximately 2.7 seconds.

## Square Root and Cube Root Functions

We begin with the square root function:

$$f(x)=\sqrt{x}$$

We know that the square root is not a real number when the radicand x is negative. Therefore, we conclude that the domain consists of all real numbers greater than or equal to 0. Here we choose 0 and some positive values for x, calculate the corresponding y-values, and plot the resulting ordered pairs.

After plotting the points, we can then sketch the graph of the square root function.

Example $$\PageIndex{11}$$

Given the function $$f(x)=\sqrt{x+2}$$, find f(−2), f(2), and f(6).

Solution:

Replace x with each of the given values.

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

\begin{aligned} f(\color{OliveGreen}{-2}\color{black}{)} &=\sqrt{\color{OliveGreen}{-2}\color{black}{+}2}=\sqrt{0}=0 \\ f(\color{OliveGreen}{2}\color{black}{)} &=\sqrt{\color{OliveGreen}{2}\color{black}{+}2}=\sqrt{4}=2 \\ f(\color{OliveGreen}{6}\color{black}{)} &=\sqrt{\color{OliveGreen}{6}\color{black}{+}2}=\sqrt{8}=\sqrt{4 \cdot 2}=2 \sqrt{2} \end{aligned}

$$f(−2)=0, f(2)=2$$, and $$f(6)=2\sqrt{2}$$

Next, consider the cube root function:

$$f(x)=\sqrt[3]{x}$$

Since the cube root could be either negative or positive, we conclude that the domain consists of all real numbers. For completeness, choose some positive and negative values for x, as well as 0, and then calculate the corresponding y-values

Plot the points and sketch the graph of the cube root function.

Example $$\PageIndex{12}$$

Given the function $$g(x)=\sqrt[3]{x-1}$$, find g(−7), g(0), and g(55).

Solution:

Replace x with each of the given values.

$$g(x)=\sqrt[3]{x-1}$$

\begin{aligned} g(\color{OliveGreen}{-7}\color{black}{)} &=\sqrt[3]{\color{OliveGreen}{-7}\color{black}{-}1}=\sqrt[3]{-8}=\sqrt[3]{(-2)^{3}}=-2 \\ g(\color{OliveGreen}{0}\color{black}{)} &=\sqrt[3]{\color{OliveGreen}{0}\color{black}{-}1}=\sqrt[3]{-1}=\sqrt[3]{(-1)^{3}}=-1 \\ g(\color{OliveGreen}{55}\color{black}{)} &=\sqrt[3]{\color{OliveGreen}{55}\color{black}{-}1}=\sqrt[3]{54}=\sqrt[3]{27 \cdot 2}=\sqrt[3]{3^{3} \cdot 2}=3 \sqrt[3]{2} \end{aligned}

$$g(−7)=−2, g(0)=−1$$, and $$g(55)=3\sqrt[3]{2}$$

## Key Takeaways

• In beginning algebra, we typically assume that all variable expressions within the radical are positive. This allows us to focus on simplifying radicals without the technical issues associated with the principal nth root.
• To simplify radical expressions, look for factors of the radicand with powers that match the index. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property $$\sqrt[n]{a^{n}}=a$$, where $$a$$ is positive.

Exercise $$\PageIndex{2}$$ simplifying radical expressions

Simplify. (Assume all variables represent positive numbers.)

1. $$\sqrt{36a^{2}}$$
2. $$\sqrt{121b^{2}}$$
3. $$\sqrt{x^{2}y^{2}}$$
4. $$\sqrt{25x^{2}y^{2}z^{2}}$$
5. $$\sqrt{180x^{3}}$$
6. $$\sqrt{150y^{3}}$$
7. $$\sqrt{49a^{3}b^{2}}$$
8. $$\sqrt{4a^{4}b^{3}c}$$
9. $$\sqrt{45x^{5}y^{3}}$$
10. $$\sqrt{50x^{6}y^{4}}$$
11. $$\sqrt{64r^{2}s^{6}t^{5}}$$
12. $$\sqrt{144r^{8}s^{6}t^{2}}$$
13. $$\sqrt{(x + 1 )^{2}}$$
14. $$\sqrt{( 2 x + 3 )^{2}}$$
15. $$\sqrt{4 ( 3 x − 1 )^{2}}$$
16. $$\sqrt{9 ( 2 x + 3 )^{2}}$$
17. $$\sqrt{9x^{3}5y^{2}}$$
18. $$\sqrt{4x^{5}9y^{4}}$$
19. $$\sqrt{m^{7}36 n^{4}}$$
20. $$\sqrt{147 m^{9}n^{6}}$$
21. $$\sqrt{2 r^{2}s^{5}25t^{4}}$$
22. $$\sqrt{36 r^{5} s^{2} t^{6}}$$
23. $$\sqrt[3]{27 a^{ 3}}$$
24. $$\sqrt[3]{125 b^{3}}$$
25. $$\sqrt[3]{250 x^{4}y^{3}}$$
26. $$\sqrt[3]{162 a^{3} b^{5}}$$
27. $$\sqrt[3]{64 x^{3}y^{6} z^{9}}$$
28. $$\sqrt[3]{216 x^{12}y^{3}}$$
29. $$\sqrt[3]{8 x^{3}y^{4}}$$
30. $$\sqrt[3]{27 x^{5}y^{3}}$$
31. $$\sqrt[3]{a^{4} b^{5} c^{6}}$$
32. $$\sqrt[3]{a^{7} b^{5} c^{3}}$$
33. $$\sqrt[3]{8 x^{4}27y^{3}}$$
34. $$\sqrt[3]{x^{5}125y^{6}}$$
35. $$\sqrt[3]{360 r^{5} s^{12} t^{13}}$$
36. $$\sqrt[3]{540r^{3}s^{2}t^{9}}$$
37. $$\sqrt[4]{81x^{4}}$$
38. $$\sqrt[4]{x^{4}y^{4}}$$
39. $$\sqrt[4]{16x^{4}y^{8}}$$
40. $$\sqrt[4]{81x^{12}y^{4}}$$
41. $$\sqrt[4]{a^{4}b^{5}c^{6}}$$
42. $$\sqrt[4]{54a^{6}c^{8}}$$
43. $$\sqrt[4]{128x^{6}}$$
44. $$\sqrt[4]{243y^{7}}$$
45. $$\sqrt[5]{32m^{10}n^{5}}$$
46. $$\sqrt[5]{37m^{9}n^{10}}$$
47. $$-3\sqrt{4x^{2}}$$
48. $$7\sqrt{9y^{2}}$$
49. $$-5\sqrt{x4x^{2}y}$$
50. $$−3\sqrt{y^{16}x^{3}y^{2}}$$
51. $$12ab\sqrt{a^{5}b^{3}}$$
52. $$6a^{2}b^{9}\sqrt{a^{7}b^{2}}$$
53. $$2x\sqrt[3]{8x^{6}}$$
54. $$−5x^{2}\sqrt[3]{27x^{3}}$$
55. $$2ab\sqrt[3]{−8a^{4}b^{5}}$$
56. $$5a^{2}b\sqrt[3]{−27a^{3}b^{3}}$$

1. $$6a$$

3. $$xy$$

5. $$6x \sqrt{5x}$$

7. $$7ab\sqrt{a}$$

9. $$3x^{2}y\sqrt{5xy}$$

11. $$8rs^{3}t^{2}\sqrt{t}$$

13. $$x+1$$

15. $$2(3x−1)$$

17. $$3xy\sqrt{5x}$$

19. $$6n^{2}m^{3}\sqrt{m}$$

21. $$5\sqrt{2s}rt^{2}s^{2}$$

23. $$3a$$

25. $$5xy\sqrt[3]{2x}$$

27. $$4xy^{2}z^{3}$$

29. $$2xy\sqrt[3]{y}$$

31. $$abc^{2}\sqrt[3]{ab^{2}}$$

33. $$6yx\sqrt[3]{x}$$

35. $$2rs^{4}t^{4}\sqrt[3]{45r^{2}t}$$

37. $$3x$$

39. $$2xy^{2}$$

41. $$abc\sqrt[4]{bc^{2}}$$

43. $$2x\sqrt[4]{8x^{2}}$$

45. $$2^{m}2^{n}$$

47. $$−6x$$

49. $$−10 x\sqrt{xy}$$

51. $$12 a^{3} b^{2}\sqrt{ab}$$

53. $$4x^{3}$$

55. $$− 4 a^{ 2} b^{ 2}\sqrt[3]{ab^{2}}$$

Exercise $$\PageIndex{3}$$ simplifying radical expressions

Rewrite the following as a radical expression with coefficient 1.

1. $$5\sqrt{2x}$$
2. $$2\sqrt{3y}$$
3. $$2x\sqrt{3}$$
4. $$3y\sqrt{2}$$
5. $$ab\sqrt{10a}$$
6. $$2ab\sqrt{2a}$$
7. $$m^{2}n\sqrt{mn}$$
8. $$2m^{2}n^{3}\sqrt{3n}$$
9. $$5\sqrt[3]{2x}$$
10. $$3\sqrt[3]{5y}$$
11. $$2x\sqrt[3]{3}$$
12. $$3y\sqrt[3]{2}$$

1. $$\sqrt{50 x}$$

3. $$\sqrt{12 x^{2}}$$

5. $$\sqrt{10 a^{3} b^{2}}$$

7. $$\sqrt{m^{5} n^{3}}$$

9. $$\sqrt[3]{250 x}$$

11. $$\sqrt[3]{24 x^{3}}$$

Exercise $$\PageIndex{4}$$ simplifying radical expressions

Assume that the variable could represent any real number and then simplify.

1. $$\sqrt{4x^{2}}$$
2. $$\sqrt{25y^{2}}$$
3. $$\sqrt[3]{8y^{3}}$$
4. $$\sqrt[3]{125a^{3}}$$
5. $$\sqrt[4]{64x^{4}}$$
6. $$\sqrt[4]{81y^{4}}$$
7. $$\sqrt{36a^{4}}$$
8. $$\sqrt{100a^{8}}$$
9. $$\sqrt{4a^{6}}$$
10. $$\sqrt{a^{10}}$$
11. $$\sqrt{18a^{4}b^{5}}$$
12. $$\sqrt{48a^{5}b^{3}}$$
13. $$\sqrt[6]{128x^{6}y^{8}}$$
14. $$\sqrt[6]{a^{6}b^{7}c^{8}}$$

1. $$2 | x |$$

3. $$2y$$

5. $$2 | x |$$

7. $$6 a^{2}$$

9. $$2| a^{3}|$$

11. $$3 a^{2} b^{2}\sqrt{2 b}$$

13. $$2 |xy |\sqrt[6]{ 2y^{2}}$$

Exercise $$\PageIndex{5}$$ formulas involving radicals

The y -intercepts for any graph will have the form (0, y), where y is a real number. Therefore, to find y -intercepts, set x = 0 and solve for y. Find the y -intercepts for the following.

1. $$y=\sqrt{x+4}−1$$
2. $$y=\sqrt{x+1}−3$$
3. $$y=\sqrt[3]{x−1}+2$$
4. $$y=\sqrt[3]{x+1}−3$$

1. $$(0, 1)$$

3. $$(0, 1)$$

Exercise $$\PageIndex{6}$$ formulas involving radicals

Use the distance formula to calculate the distance between the given two points.

1. $$(5, −7)$$ and $$(3, −8)$$
2. $$(−9, 7)$$ and $$(−8, 4)$$
3. $$(−3, −4)$$ and $$(3, −6)$$
4. $$(−5, −2)$$ and $$(1, −6)$$
5. $$(−1, 1)$$ and $$(−4, 10)$$
6. $$(8, −3)$$ and $$(2, −12)$$

1. $$\sqrt{5}$$

3. $$2\sqrt{10}$$

5. $$3\sqrt{10}$$

Exercise $$\PageIndex{7}$$ formulas involving radicals

Factor the radicand and then simplify. (Assume that all expressions are positive.)

1. $$\sqrt{x^{2}−6x+9}$$
2. $$\sqrt{x^{2}−10x+25}$$
3. $$\sqrt{4x^{2}+12x+9}$$
4. $$\sqrt{9x^{2}+6x+1}$$
5. The speed of a vehicle before the brakes were applied can be estimated by the length of the skid marks left on the road. On dry pavement, the speed, v, in miles per hour can be estimated by the formula $$v=\sqrt{5d}$$, where d represents the length of the skid marks in feet. Estimate the speed of a vehicle before applying the brakes on dry pavement if the skid marks left behind measure 36 feet.
6. The radius, r, of a sphere can be calculated using the formula $$r=\sqrt[3]{\frac{3 V}{4 \pi}}$$, where V represents the sphere’s volume. What is the radius of a sphere if the volume is $$36π$$ cubic centimeters?

1. $$x − 3$$

3. $$2 x + 3$$

5. $$30$$ miles per hour

Exercise $$\PageIndex{8}$$ formulas involving radicals

The period, T, of a pendulum in seconds is given by the formula

$$T=2 \pi \sqrt{\frac{L}{32}}$$

where L represents the length in feet. Calculate the period, given the following lengths. Give the exact value and the approximate value rounded off to the nearest tenth of a second.

1. 8 feet
2. 32 feet
3. 1/2 foot
4. 1/8 foot

1. $$π≈3.1$$ seconds

3. $$\frac{π}{4} ≈0.8$$ seconds

Exercise $$\PageIndex{9}$$ formulas involving radicals

The time, t, in seconds that an object is in free fall is given by the formula

$$s=16\cdot t^{2}$$

where s represents the distance it has fallen in feet. Calculate the time it takes an object to fall, given the following distances. Give the exact value and the approximate value rounded off to the nearest tenth of a second.

1. 48 feet
2. 80 feet
3. 192 feet
4. 288 feet

1. $$\sqrt{3} ≈1.7$$ seconds

3. $$2\sqrt{3} \approx 3.5$$ seconds

Exercise $$\PageIndex{10}$$ radical functions

Given the function, calculate the following.

1. $$f(x)=\sqrt{x−1}$$, find f(1), f(2), and f(5)
2. $$f(x)=\sqrt{x+5}$$, find f(−5), f(−1), and f(20)
3. $$f(x)=\sqrt{x}+3$$, find f(0), f(1), and f(16)
4. $$f(x)=\sqrt{x}−5$$, find f(0), f(1), and f(25)
5. $$g(x)=\sqrt[3]{x}$$, find g(−1), g(0), and g(1)
6. $$g(x)=\sqrt[3]{x+7}$$, find g(−15), g(−7), and g(20)
7. $$g(x)=\sqrt[3]{x}−2$$, find g(−1), g(0), and g(8)
8. $$g(x)=\(\sqrt[3]{x−1}+2$$, find g(0), g(2), and g(9)

1. $$f(1)=0, f(2)=1$$, and $$f(5)=2$$

3. $$f(0)=3, f(1)=4$$, and $$f(16)=7$$

5. $$g(−1)=−1, g(0)=0$$, and $$g(1)=1$$

7. $$g(−1)=−3, g(0)=−2$$, and $$g(8)=0$$

Exercise $$\PageIndex{11}$$ radical functions

For each function, fill in the table.

1. $$f(x)=\sqrt{x+1}$$

Figure $$\PageIndex{7}$$
2. $$f(x)=\sqrt{x−2}$$

Figure $$\PageIndex{8}$$
3. $$f(x)=\sqrt[3]{x}+1$$

Figure $$\PageIndex{9}$$
4. $$f(x)=\sqrt[3]{x+2}$$

Figure $$\PageIndex{10}$$

1.

3.

Exercise $$\PageIndex{12}$$ discussion board

1. Give a value for x such that $$\sqrt{x^{2}}≠x$$. Explain why it is important to assume that the variables represent positive numbers.
2. Research and discuss the accomplishments of Christoph Rudolff. What is he credited for?
3. Research and discuss the methods used for calculating square roots before the common use of electronic calculators.
4. What is a surd, and where does the word come from?