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3.4 Solving Rational Equations and Radical Equations

Solve rational equations.
Solve radical equations.

Rational Equations

Equations containing rational expressions are called rational equations. Solving such equations involves multiplying on both sides by the least common denominator (LCD) of all the rational expressions to clear the equation of fractions.

Example 1

Solve: x−83+x−32=0 $\frac{x - 8}{3} + \frac{x - 3}{2} = 0$ .

The zero of the function is 5. Thus the solution of the equation is 5.

Now Try Exercise 3.

Caution!

Clearing fractions is a valid procedure when solving rational equations but not when adding, subtracting, multiplying, or dividing rational expressions. A rational expression may have operation signs but it will have no equals sign. A rational equation always has an equals sign. For example, x−83+x−32 $\frac{x - 8}{3} + \frac{x - 3}{2}$ is a rational expression but x−83+x−32=0 $\frac{x - 8}{3} + \frac{x - 3}{2} = 0$ is a rational equation.

To simplify the rational expression x−83+x−32 $\frac{x - 8}{3} + \frac{x - 3}{2}$ , we first find the LCD and write each fraction with that denominator. The final result is usually a rational expression.

To solve the rational equation x−83+x−32=0 $\frac{x - 8}{3} + \frac{x - 3}{2} = 0$ , we first multiply by the LCD on both sides to clear fractions. The final result is one or more numbers. As we will see in Example 2, these numbers must be checked in the original equation.

When we use the multiplication principle to multiply (or divide) on both sides of an equation by an expression with a variable, we might not obtain an equivalent equation. We must check the possible solutions obtained in this manner by substituting them in the original equation. The next example illustrates this.

Example 2

Solve: x2x−3=9x−3 $\frac{x^{2}}{x - 3} = \frac{9}{x - 3}$ .

Solution

The LCD is x−3 $x - 3$ .

(x - 3) \cdot x 2 x - 3 x 2 = = (x - 3) \cdot 9 x - 3 9 x = - 3 o r x = 3 U s i n g t h e p r i n c i p l e o f s q u a r e r o o t s

$\begin{array}{l} \begin{array}{rcl} (x - 3) \cdot \frac{x^{2}}{x - 3} & = & (x - 3) \cdot \frac{9}{x - 3} \\ x^{2} & = & 9 \end{array} \\ \begin{array}{l} x = - 3 & or & x = 3 & Using the principle of square roots \end{array} \end{array}$

The possible solutions are −3 and 3. We check.

Check:

For −3

For 3

The number −3 checks, so it is a solution. Since division by 0 is not defined, 3 is not a solution. Note that 3 is not in the domain of either x2/(x−3) $x^{2} / (x - 3)$ or 9/(x−3) $9 / (x - 3)$ .

Now Try Exercise 9.

Example 3

Solve: 23x+6+1x2−4=4x−2 $\frac{2}{3 x + 6} + \frac{1}{x^{2} −4} = \frac{4}{x - 2}$ .

Solution

We first factor the denominators in order to determine the LCD:

23(x+2)+1(x+2)(x−2)3(x+2)(x−2)(23(x+2)+1(x+2)(x−2))2(x−2)+32x−4+32x−1−10xx=======4x−2The LCD is 3(x+2)(x−2).3(x+2)(x−2)⋅4x−2Multiplying by the LCD to clear fractions3⋅4(x−2)+312x+2412x+2425−52.

$\begin{array}{rcl} \frac{2}{3 (x + 2)} + \frac{1}{(x + 2) (x - 2)} & = & \frac{4}{x - 2} \begin{array}{l} \begin{array}{l} The LCD is \\ 3 (x + 2) (x - 2) . \end{array} \end{array} \\ 3 (x + 2) (x - 2) (\frac{2}{3 (x + 2)} + \frac{1}{(x + 2) (x - 2)}) & = & 3 (x + 2) (x - 2) \cdot \frac{4}{x - 2} \\ Multiplying by the LCD to clear fractions \\ 2 (x - 2) + 3 & = & 3 \cdot 4 (x - 2) + 3 \\ 2 x - 4 + 3 & = & 12 x + 24 \\ 2 x - 1 & = & 12 x + 24 \\ −10 x & = & 25 \\ x & = & - \frac{5}{2} . \end{array}$

The possible solution is −52 $- \frac{5}{2}$ . This number checks. It is the solution.

Now Try Exercise 21.

Radical Equations

A radical equation is an equation in which variables appear in one or more radicands. For example,

2 x - 5 - - - - - \sqrt - x - 3 - - - - - \sqrt = 1

$\sqrt{2 x - 5} - \sqrt{x - 3} = 1$

is a radical equation. The following principle is used to solve such equations.

Example 4

Solve: 3x+1−−−−−√=4 $\sqrt{3 x + 1} = 4$ .

Solution

We have

3 x + 1 - - - - - \sqrt (3 x + 1 - - - - - \sqrt) 2 3 x + 1 3 x x = = = = = 4421615 5. U s i n g t h e p r i n c i p l e o f p o w e r s; s q u a r i n g b o t h s i d e s

$\begin{array}{rcll} \sqrt{3 x + 1} & = & 4 \\ {(\sqrt{3 x + 1})}^{2} & = & 4^{2} & Using the principle of powers; squaring both sides \\ 3 x + 1 & = & 16 \\ 3 x & = & 15 \\ x & = & 5. \end{array}$

Check:

The solution is 5.

Now Try Exercise 31.

In Example 4, the radical was isolated on one side of the equation. If this had not been the case, our first step would have been to isolate the radical. We do so in the next example.

Example 5

Solve: 5+x+7−−−−−√=x $5 + \sqrt{x + 7} = x$ .

Now Try Exercise 55.

When we raise both sides of an equation to an even power, the resulting equation can have solutions that the original equation does not. This is because the converse of the principle of powers is not necessarily true. That is, if an=bn $a^{n} = b^{n}$ is true, we do not know that a=b $a = b$ is true. For example, ${(−2)}^{2} = 2^{2}$ , but $−2 \neq 2$ . Thus, as we saw in Example 5, it is necessary to check the possible solutions in the original equation when the principle of powers is used to raise both sides of an equation to an even power.

When a radical equation has two radical terms on one side, we isolate one of them and then use the principle of powers. If, after doing so, a radical term remains, we repeat these steps.

Example 6

Solve: $\sqrt{x - 3} + \sqrt{x + 5} = 4$ .

Solution

We have

$\begin{array}{rcll} \sqrt{x - 3} & = & 4 - \sqrt{x + 5} & Isolating one radical \\ {(\sqrt{x - 3})}^{2} & = & {(4 - \sqrt{x + 5})}^{2} & Using the principle of \\ powers; squaring both sides \\ x - 3 & = & 16 - 8 \sqrt{x + 5} + (x + 5) \\ x - 3 & = & 21 - 8 \sqrt{x + 5} + x & Combining like terms \\ −24 & = & −8 \sqrt{x + 5} & Isolating the remaining \\ radical; subtracting x and \\ 21 on both sides \\ 3 & = & \sqrt{x + 5} & Dividing by - 8 on both sides \\ 3^{2} & = & {(\sqrt{x + 5})}^{2} & Using the principle of \\ powers; squaring both sides \\ 9 & = & x + 5 \\ 4 & = & x . & Subtracting 5 on both sides \end{array}$

The number 4 checks. It is the solution.

Now Try Exercise 65.

3.4 Exercise Set

Solve.

1. $\frac{1}{4} + \frac{1}{5} = \frac{1}{t}$
2. $\frac{1}{3} - \frac{5}{6} = \frac{1}{x}$
3. $\frac{x + 2}{4} - \frac{x - 1}{5} = 15$
4. $\frac{t + 1}{3} - \frac{t - 1}{2} = 1$
5. $\frac{1}{2} + \frac{2}{x} = \frac{1}{3} + \frac{3}{x}$
6. $\frac{1}{t} + \frac{1}{2 t} + \frac{1}{3 t} = 5$
7. $\frac{5}{3 x + 2} = \frac{3}{2 x}$
8. $\frac{2}{x - 1} = \frac{3}{x + 2}$
9. $\frac{y^{2}}{y + 4} = \frac{16}{y + 4}$
10. $\frac{49}{w - 7} = \frac{w^{2}}{w - 7}$
11. $x + \frac{6}{x} = 5$
12. $x - \frac{12}{x} = 1$
13. $\frac{6}{y + 3} + \frac{2}{y} = \frac{5 y - 3}{y^{2} - 9}$
14. $\frac{3}{m + 2} + \frac{2}{m} = \frac{4 m - 4}{m^{2} - 4}$
15. $\frac{2 x}{x - 1} = \frac{5}{x - 3}$
16. $\frac{2 x}{x + 7} = \frac{5}{x + 1}$
17. $\frac{2}{x + 5} + \frac{1}{x - 5} = \frac{16}{x^{2} - 25}$
18. $\frac{2}{x^{2} - 9} + \frac{5}{x - 3} = \frac{3}{x + 3}$
19. $\frac{3 x}{x + 2} + \frac{6}{x} = \frac{12}{x^{2} + 2 x}$
20. $\frac{3 y + 5}{y^{2} + 5 y} + \frac{y + 4}{y + 5} = \frac{y + 1}{y}$
21. $\frac{1}{5 x + 20} - \frac{1}{x^{2} - 16} = \frac{3}{x - 4}$
22. $\frac{1}{4 x + 12} - \frac{1}{x^{2} - 9} = \frac{5}{x - 3}$
23. $\frac{2}{5 x + 5} - \frac{3}{x^{2} - 1} = \frac{4}{x - 1}$
24. $\frac{1}{3 x + 6} - \frac{1}{x^{2} - 4} = \frac{3}{x - 2}$
25. $\frac{8}{x^{2} - 2 x + 4} = \frac{x}{x + 2} + \frac{24}{x^{3} + 8}$
26. $\frac{18}{x^{2} - 3 x + 9} - \frac{x}{x + 3} = \frac{81}{x^{3} + 27}$
27. $\frac{x}{x - 4} - \frac{4}{x + 4} = \frac{32}{x^{2} - 16}$
28. $\frac{x}{x - 1} - \frac{1}{x + 1} = \frac{2}{x^{2} - 1}$
29. $\frac{1}{x - 6} - \frac{1}{x} = \frac{6}{x^{2} - 6 x}$
30. $\frac{1}{x - 15} - \frac{1}{x} = \frac{15}{x^{2} - 15 x}$
31. $\sqrt{3 x - 4} = 1$
32. $\sqrt{4 x + 1} = 3$
33. $\sqrt{2 x - 5} = 2$
34. $\sqrt{3 x + 2} = 6$
35. $\sqrt{7 - x} = 2$
36. $\sqrt{5 - x} = 1$
37. $\sqrt{1 - 2 x} = 3$
38. $\sqrt{2 - 7 x} = 2$
39. $\sqrt[3]{5 x - 2} = - 3$
40. $\sqrt[3]{2 x + 1} = - 5$
41. $\sqrt[4]{x^{2} - 1} = 1$
42. $\sqrt[5]{3 x + 4} = 2$
43. $\sqrt{y - 1} + 4 = 0$
44. $\sqrt{m + 1} - 5 = 8$
45. $\sqrt{b + 3} - 2 = 1$
46. $\sqrt{x - 4} + 1 = 5$
47. $\sqrt{z + 2} + 3 = 4$
48. $\sqrt{y - 5} - 2 = 3$
49. $\sqrt{2 x + 1} - 3 = 3$
50. $\sqrt{3 x - 1} + 2 = 7$
51. $\sqrt{2 - x} - 4 = 6$
52. $\sqrt{5 - x} + 2 = 8$
53. $\sqrt[3]{6 x + 9} + 8 = 5$
54. $\sqrt[5]{2 x - 3} - 1 = 1$
55. $\sqrt{x + 4} + 2 = x$
56. $\sqrt{x + 1} + 1 = x$
57. $\sqrt{x - 3} + 5 = x$
58. $\sqrt{x + 3} - 1 = x$
59. $\sqrt{x + 7} = x + 1$
60. $\sqrt{6 x + 7} = x + 2$
61. $\sqrt{3 x + 3} = x + 1$
62. $\sqrt{2 x + 5} = x - 5$
63. $\sqrt{5 x + 1} = x - 1$
64. $\sqrt{7 x + 4} = x + 2$
65. $\sqrt{x - 3} + \sqrt{x + 2} = 5$
66. $\sqrt{x} - \sqrt{x - 5} = 1$
67. $\sqrt{3 x - 5} + \sqrt{2 x + 3} + 1 = 0$
68. $\sqrt{2 m - 3} = \sqrt{m + 7} - 2$
69. $\sqrt{x} - \sqrt{3 x - 3} = 1$
70. $\sqrt{2 x + 1} - \sqrt{x} = 1$
71. $\sqrt{2 y - 5} - \sqrt{y - 3} = 1$
72. $\sqrt{4 p + 5} + \sqrt{p + 5} = 3$
73. $\sqrt{y + 4} - \sqrt{y - 1} = 1$
74. $\sqrt{y + 7} + \sqrt{y + 16} = 9$
75. $\sqrt{x + 5} + \sqrt{x + 2} = 3$
76. $\sqrt{6 x + 6} = 5 + \sqrt{21 - 4 x}$
77. $x^{1 / 3} = - 2$
78. $t^{1 / 5} = 2$
79. $t^{1 / 4} = 3$
80. $m^{1 / 2} = - 7$

Solve.

81. $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ , for $T_{1}$ (A chemistry formula for gases)
82. $\frac{1}{F} = \frac{1}{m} + \frac{1}{p}$ , for $F$ (A formula from optics)
83. $W = \sqrt{\frac{1}{L C}}$ , for $C$ (An electricity formula)
84. $s = \sqrt{\frac{A}{6}}$ , for $A$ (A geometry formula)
85. $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ , for $R_{2}$ (A formula for resistance)
86. $\frac{1}{t} = \frac{1}{a} + \frac{1}{b}$ , for $t$ (A formula for work rate)
87. $I = \sqrt{\frac{A}{P}} - 1$ , for $P$ (A compound-interest formula)
88. $T = 2 π \sqrt{\frac{1}{g}}$ , for $g$ (A pendulum formula)
89. $\frac{1}{F} = \frac{1}{m} + \frac{1}{p}$ , for $p$ (A formula from optics)
90. $\frac{V^{2}}{R^{2}} = \frac{2 g}{R + h}$ , for $h$ (A formula for escape velocity)

Skill Maintenance

Find the zero of the function. [1.5]

91. $f (x) = 15 - 2 x$
92. $f (x) = - 3 x + 9$

Solve. [1.5]

93. Pork Production. Together, China and the United States, the top two pork producers worldwide, produced 64,308,000 metric tons of pork in 2013. China produced 1,260,000 metric tons more than five times the number of metric tons produced by the United States. (Source: United Nations Food and Agriculture Organization) How many metric tons of pork did each country produce in 2013?
94. Sports Injuries. In 2012 in the United States, there were 172,470 injuries among soccer players ages 19 and under. This was about 44% more than the number of injuries among baseball players ages 19 and under. (Source: Safe Kids Worldwide, based on hospital ER reports, 2012) How many baseball players ages 19 and under were injured in 2012?

Synthesis

Solve.

95. ${(x - 3)}^{2 / 3} = 2$
96. $\frac{x + 3}{x + 2} - \frac{x + 4}{x + 3} = \frac{x + 5}{x + 4} - \frac{x + 6}{x + 5}$
97. $\sqrt{x + 5} + 1 = \frac{6}{\sqrt{x + 5}}$
98. $\sqrt{15 + \sqrt{2 x + 80}} = 5$
99. $x^{2 / 3} = x$

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Table of Contents for 3.4 Solving Rational Equations and Radical Equations

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Table of Contents for
3.4 Solving Rational Equations and Radical Equations