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Section 1.5 Solving Other Types of Equations

Before Starting this Section, Review

1 Factoring (Section P.4 )
2 Rational expressions (Section P.5 )
3 Rational exponents and radicals (Section P.6 , page 61)
4 Quadratic equations (Sections 1.3 and 1.4)

Objectives

1 Solve equations by factoring.
2 Solve rational equations.
3 Solve equations involving radicals.
4 Solve equations with rational exponents.
5 Solve equations that are quadratic in form.

Time Dilation

Do you ever feel as though time moves really quickly or really slowly? For example, don’t the hours fly by when you are hanging out with your best friend and the seconds drag on endlessly when a dentist is working on your teeth? But you cannot really speed up or slow time down, right? It always “flows” at the same rate.

Einstein said “no.” He asserted in his “Special Theory of Relativity” that time is relative. It speeds up or slows down depending on how fast one object is moving relative to another. He concluded that the closer we come to traveling at the speed of light, the more time will slow down for us relative to someone not moving. He called this slowing of time due to motion time dilation. Time dilation can be quantified by the equation

$t_{0} = t \sqrt{1 - \frac{v^{2}}{c^{2}}},$

where $t_{0} = the$ “proper time” measured in the moving frame of reference,

$\begin{array}{rcll} t & = & the time observed in the fixed frame of reference, \\ v & = & the speed of the moving frame of reference, and \\ c & = & the speed of light . \end{array}$

We investigate this effect in Example 11.

Albert Einstein 1879–1955

Einstein began his schooling in Munich, Germany. Einstein renounced his German citizenship in 1896 and entered the Swiss Federal Polytechnic School in Zurich to be trained as a teacher in physics and mathematics. Unable to find a teaching job, he accepted a position as a technical assistant in the Swiss patent office. In 1905, Einstein earned his doctorate from the University of Zurich.

While at the patent office, he produced much of his remarkable work on the special theory of relativity. In 1909, he was appointed Professor Extraordinary at Zurich. In 1914, he was appointed professor at the University of Berlin. He immigrated to America to take the position of professor of physics at Princeton University. By 1949, Einstein was unwell. He drew up his will in 1950.

He left his scientific papers to the Hebrew University in Jerusalem. In 1952, the Israeli government offered the post of second president to Einstein. He politely declined the offer.

One week before his death, Einstein agreed to let Bertrand Russell place his name on a manifesto urging all nations to give up nuclear weapons.

Solving Equations by Factoring

1 Solve equations by factoring.

The factoring method used in solving quadratic equations can also be used to solve certain nonquadratic equations. Specifically, we use the following strategy for solving those equations that can be expressed in factored form with 0 on one side.

Procedure in Action

Example 1 Solving Equations by Factoring

OBJECTIVE	EXAMPLE
Use factoring techniques to solve an equation.	Solve $x^{3} + 3 x^{2} = 4 x + 12 .$
Step 1 Make one side zero. Move all nonzero terms in the equation to one side (say, the left side) so that the other side (the right side) is 0.	$\begin{array}{rcll} x^{3} + 3 x^{2} - 4 x - 12 = 0 & \begin{array}{l} Subtract 4 x + 12 \\ from both sides . \end{array} \end{array}$
Step 2 Factor the left side.	$\begin{array}{rcll} x^{2} (x + 3) - 4 (x + 3) & = & 0 & Group for factoring . \\ (x + 3) (x^{2} - 4) & = & 0 & Factor out x + 3 . \end{array}$
Step 3 Use the zero-product property. Set each factor in Step 2 equal to 0 and then solve the resulting equations.	$\begin{array}{l} \begin{aligned} (x + 3) (x + 2) (x - 2) = 0 & Factor x^{2} - 4 . \end{aligned} \\ \begin{array}{rcllrcllrcl} x + 3 & = & 0 & or & x + 2 & = & 0 & or & x - 2 & = & 0 \\ x & = & - 3 & or & x & = & - 2 & or & x & = & 2 \end{array} \end{array}$
Step 4 Check your solutions.	Check: Check the solutions in the original equation. For $\begin{array}{lrcll} x = - 3, {(- 3)}^{3} & + 3 {(- 3)}^{2} & \overset{?}{=} & 4 (- 3) + 12 \\ - 27 + 27 & = & - 12 + 12 ✓ \end{array}$ You should also verify that $x = - 2$ and $x = 2$ are solutions of $x^{3} + 3 x^{2} = 4 x + 12 .$ The solution set is ${- 3, - 2, 2} .$

Practice Problem 1

Solve: $x^{3} + 2 x^{2} - x - 2 = 0$

Example 2 Solving an Equation by Factoring

Solve by factoring: $x^{4} = 9 x^{2}$

Solution

Step 1 Move all terms to the left side.

$\begin{array}{rcll} x^{4} & = & 9 x^{2} & Original equation \\ x^{4} - 9 x^{2} & = & 0 & Subtract 9 x^{2} from both sides . \end{array}$
Step 2 Factor.

$\begin{array}{rcll} x^{2} (x^{2} - 9) & = & 0 & x^{4} - 9 x^{2} = x^{2} (x^{2} - 9) \\ x^{2} (x + 3) (x - 3) & = & 0 & x^{2} - 9 = (x + 3) (x - 3) \end{array}$
Step 3 Set each factor equal to zero.

$\begin{aligned} x^{2} = 0 & or & x + 3 & = & 0 & or & x - 3 = 0 & Zero-product property \\ x = 0 & or & x & = & - 3 & or & x = 3 & Solve each equation for x . \end{aligned}$
Step 4 Check:

$\begin{array}{rclrclrcl} Let x & = & 0 & Let x & = & - 3 & Let x & = & 3 \\ {(0)}^{4} & \overset{?}{=} & 9 {(0)}^{2} & {(- 3)}^{4} & \overset{?}{=} & 9 {(- 3)}^{2} & {(3)}^{4} & \overset{?}{=} & 9 {(3)}^{2} \\ 0 & = & 0 ✓ & 81 & = & 81 ✓ & 81 & = & 81 ✓ \end{array}$

Thus, the solution set of the equation is ${- 3, 0, 3} .$ Note that 0 is a root of multiplicity 2.

Practice Problem 2

Solve: $x^{4} = 4 x^{2}$

Warning

A common error when solving an equation such as $x^{4} = 9 x^{2}$ in Example 2 is to divide both sides by $x^{2},$ obtaining $x^{2} = 9$ so that $\pm 3$ are the two solutions. This leads to the loss of the solution 0. Remember, we can divide only by quantities that are not 0.

Rational Equations

2 Solve rational equations.

Recall that if at least one algebraic expression with the variable in the denominator appears in an equation, then the equation is a rational equation.

When we multiply a rational equation by an expression containing the variable, we may introduce a solution that satisfies the new equation but does not satisfy the original equation. Such a solution is called an extraneous solution or extraneous root. So whenever we multiply an equation by an expression containing the variable, we must check all possible solutions obtained to reject extraneous solutions (if any).

Example 3 Solving a Rational Equation

Solve: $\frac{1}{6} + \frac{1}{x + 1} = \frac{1}{x}$

Solution

Step 1 Find the LCD. The LCD from the denominators

$6, x + 1, and x is 6 x (x + 1) .$
Step 2 Multiply both sides of the equation by the LCD and then make the right side 0.
Step 3 Factor.

$\begin{array}{rcll} x^{2} + x - 6 & = & 0 \\ (x + 3) (x - 2) & = & 0 & Factor . \end{array}$
Step 4 Set each factor equal to zero.

$\begin{array}{rcllr} x + 3 & = & 0 & or & x - 2 = 0 \\ x & = & - 3 & or & x = 2 \end{array}$
Step 5 Check: You should check the solutions, $- 3$ and 2, in the original equation. The solution set is ${- 3, 2} .$

Practice Problem 3

Solve: $\frac{1}{x} - \frac{12}{5 x + 10} = \frac{1}{5}$

Example 4 Solving a Rational Equation with an Extraneous Solution

Solve: $\frac{x}{x - 1} - \frac{1}{x + 1} = \frac{2 x}{x^{2} - 1}$

Solution

Step 1 The LCD of the denominators $(x - 1), (x + 1), (x^{2} - 1) = (x - 1) (x + 1)$ is $(x - 1) (x + 1) .$
Step 2

$\begin{array}{l} \begin{array}{cl} (x - 1) (x + 1) [\frac{x}{x - 1} - \frac{1}{x + 1}] & \begin{array}{l} Multiply both sides by \\ the LCD . \end{array} \\ = (x - 1) (x + 1) [\frac{2 x}{x^{2} - 1}] \end{array} \\ \begin{array}{l} (x - 1) (x + 1) \frac{x}{x - 1} - (x - 1) (x + 1) \frac{1}{x + 1} & \begin{array}{l} Distributive property; \\ (x - 1) (x + 1) = x^{2} - 1 \end{array} \\ = (x^{2} - 1) \cdot \frac{2 x}{x^{2} - 1} \end{array} \\ \begin{array}{rcll} (x + 1) x - (x - 1) & = & 2 x & Simplify . \\ x^{2} + x - x + 1 - 2 x & = & 2 x - 2 x & Distributive property; add \\ - 2 x to both sides . \\ x^{2} - 2 x + 1 & = & 0 & Simplify . \end{array} \end{array}$
Step 3 $\begin{array}{l} (x - 1) (x - 1) = 0 & Factor . \end{array}$
Step 4 $\begin{array}{rcll} x - 1 = 0 & or & x = 1 & Solve for x . \end{array}$
Step 5 Check: Substitute $x = 1$ in the original equation.

Therefore, $x = 1$ is not a solution of the original equation; it is extraneous. The original equation has no solution. Its solution set is $\emptyset$ .

Practice Problem 4

Solve: $\frac{x}{x - 2} - \frac{2}{x + 2} = \frac{4 x}{x^{2} - 4}$

Equations Involving Radicals

3 Solve equations involving radicals.

If an equation involves radicals or rational exponents, the method of raising both sides to a positive integer power is often used to remove them. When this is done, the solution set of the new equation always contains the solutions of the original equation. However, in some cases, the new equation has more solutions than the original equation. For instance, consider the equation $x = 4 .$ If we square both sides, we get $x^{2} = 16 .$ Notice that the given equation, $x = 4,$ has only one solution (namely, 4), whereas the new equation, $x^{2} = 16,$ has two solutions: 4 and $- 4 .$ Extraneous solutions may be introduced whenever both sides of an equation are raised to an even power, Thus, it is essential that we check all solutions obtained after raising both sides of an equation to any even power.

Side Note

If A and B are algebraic expressions, then every solution of the equation $A = B$ is also a solution of the equation $A^{n} = B^{n},$ for any positive integer n.

However, solutions of $A^{n} = B^{n}$ are not necessarily solutions of $A = B .$

Example 5 Solving Equations Involving Radicals

Solve: $x = \sqrt{6 x - x^{3}}$

Solution

Because ${(\sqrt{a})}^{2} = a,$ we raise both sides of the equation to the second power to eliminate the radical sign. The solutions of the original equation are among the solutions of the equation

$\begin{array}{l} \begin{array}{rcll} x^{2} & = & {(\sqrt{6 x - x^{3}})}^{2} & Square both sides of the equation . \\ x^{2} & = & 6 x - x^{3} & {(\sqrt{6 x - x^{3}})}^{2} = 6 x - x^{3} \\ x^{3} + x^{2} - 6 x & = & 0 & Add x^{3} - 6 x to both sides . \\ x (x^{2} + x - 6) & = & 0 & Factor out x from each term . \\ x (x + 3) (x - 2) & = & 0 & Factor x^{2} + x - 6 . \end{array} \\ \begin{array}{llrcllrl} x = 0 & or & x + 3 & = & 0 & or & x - 2 = 0 & Zero-product property \\ x = 0 & or & x & = & - 3 & or & x = 2 & Solve each equation . \end{array} \end{array}$

Hence, the only possible solutions of $x = \sqrt{6 x - x^{3}}$ are $- 3, 0,$ and 2.

Check:

$\begin{array}{c} Let x = - 3 . & Let x = 0 . & Let x = 2 . \\ \begin{array}{rcl} - 3 & \overset{?}{=} & \sqrt{6 (- 3) - {(- 3)}^{3}} \\ - 3 & \overset{?}{=} & \sqrt{- 18 + 27} \\ - 3 & \overset{?}{=} & \sqrt{9} \\ - 3 & \neq & 3 ✕ \end{array} & \begin{array}{rcl} 0 & \overset{?}{=} & \sqrt{6 (0) - {(0)}^{3}} \\ 0 & \overset{?}{=} & \sqrt{0} \\ 0 & \overset{?}{=} & 0 ✓ \end{array} & \begin{array}{rcl} 2 & \overset{?}{=} & \sqrt{6 (2) - {(2)}^{3}} \\ 2 & \overset{?}{=} & \sqrt{12 - 8} \\ 2 & \overset{?}{=} & \sqrt{4} \\ 2 & = & 2 ✓ \end{array} \end{array}$

Thus, $- 3$ is an extraneous solution; 0 and 2 are the solutions of the equation $x = \sqrt{6 x - x^{3}} .$ The solution set is ${0, 2} .$

Practice Problem 5

Solve: $x = \sqrt{x^{3} - 6 x}$

Example 6 Solving an Equation Involving a Radical

Solve: $\sqrt{2 x + 1} + 1 = x$

Solution

Step 1 We begin by isolating the radical to one side of the equation.

$\begin{array}{rcll} \sqrt{2 x + 1} + 1 & = & x & Original equation \\ \sqrt{2 x + 1} & = & x - 1 & Subtract 1 from both sides . \end{array}$
Step 2 Next, we square both sides and simplify.

$\begin{array}{l} \begin{array}{rcll} {(\sqrt{2 x + 1})}^{2} & = & {(x - 1)}^{2} & Square both sides . \\ 2 x + 1 & = & x^{2} - 2 x + 1 & {(\sqrt{a})}^{2} = a; \\ {(A - B)}^{2} = A^{2} - 2 A B + B^{2} \end{array} \\ \begin{array}{rcll} 2 x + 1 - (2 x + 1) & = & x^{2} - 2 x + 1 - (2 x + 1) & Subtract 2 x + 1 \\ from both sides . \\ 0 & = & x^{2} - 4 x & Simplify . \\ 0 & = & x (x - 4) & Factor . \end{array} \end{array}$
Step 3 Set each factor equal to zero.

$\begin{aligned} x = 0 & or & x - 4 = 0 \\ x = 0 & or & x = 4 & Solve for x . \end{aligned}$
Step 4 Check:

$\begin{array}{c} Let x = 0 . & Let x = 4 . \\ \begin{array}{rcl} \sqrt{2 (0) + 1} + 1 & \overset{?}{=} & 0 \\ \sqrt{1} + 1 & \overset{?}{=} & 0 \\ 1 + 1 & \overset{?}{=} & 0 \\ 2 & \neq & 0 \end{array} & \begin{array}{rcl} \sqrt{2 (4) + 1} + 1 & \overset{?}{=} & 4 \\ \sqrt{9} + 1 & \overset{?}{=} & 4 \\ 3 + 1 & \overset{?}{=} & 4 \\ 4 & = & 4 ✓ \end{array} \end{array}$

Thus, 0 is an extraneous root; the only solution of the given equation is 4, and the solution set is {4}.

Practice Problem 6

Solve: $\sqrt{6 x + 4} + 2 = x$

Next, we illustrate a method for solving radical equations that contain two or more square root expressions.

Example 7 Solving an Equation Involving Two Radicals

Solve: $\sqrt{2 x - 1} - \sqrt{x - 1} = 1$

Solution

Step 1 We first isolate one of the radicals.

$\begin{array}{rcll} \sqrt{2 x - 1} - \sqrt{x - 1} & = & 1 & Original equation \\ \sqrt{2 x - 1} & = & 1 + \sqrt{x - 1} & Add \sqrt{x - 1} to both sides to \\ isolate the radical \sqrt{2 x - 1} . \end{array}$
Step 2 Square both sides of the last equation.

$\begin{array}{rcll} {(\sqrt{2 x - 1})}^{2} & = & {(1 + \sqrt{x - 1})}^{2} & Square both sides . \\ 2 x - 1 & = & 1^{2} + 2 \cdot 1 \cdot \sqrt{x - 1} + {(\sqrt{x - 1})}^{2} & {(A + B)}^{2} = A^{2} + 2 A B + B^{2} \\ 2 x - 1 & = & 1 + 2 \sqrt{x - 1} + x - 1 & Simplify; use {(\sqrt{a})}^{2} = a . \\ 2 x - 1 & = & 2 \sqrt{x - 1} + x & Simplify . \end{array}$
Step 3 The last equation still contains a radical. We repeat the process of isolating the radical expression.

$\begin{array}{rcll} x - 1 & = & 2 \sqrt{x - 1} & Subtract x from both sides and \\ simplify to isolate 2 \sqrt{x - 1} . \\ {(x - 1)}^{2} & = & {(2 \sqrt{x - 1})}^{2} & Square both sides . \\ x^{2} - 2 x + 1 & = & 4 (x - 1) & Use {(A - B)}^{2} = A^{2} - 2 A B + B^{2} \\ and {(A B)}^{2} = A^{2} B^{2} . \\ x^{2} - 2 x + 1 & = & 4 x - 4 & Distributive property \\ x^{2} - 6 x + 5 & = & 0 & Add - 4 x + 4 to both sides; simplify . \\ (x - 5) (x - 1) & = & 0 & Factor . \end{array}$
Step 4 Set each factor equal to 0.

$\begin{array}{cccl} x - 5 = 0 & or & x - 1 = 0 \\ x = 5 & or & x = 1 & Solve for x . \end{array}$
Step 5 Check: You should check that the solutions of the original equation are 1 and 5. The solution set is {1, 5}.

Practice Problem 7

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

Equations with Rational Exponents

4 Solve equations with rational exponents.

If a given equation can be expressed in the form $u^{m / n} = k,$ then we isolate u by raising both sides to the power $\frac{n}{m} (the reciprocal of \frac{m}{n}) .$ Care must be taken with even roots.

Example 8 Solving Equations with Rational Exponents

Solve.

$x^{3 / 5} = 8$
$2 {(2 x - 1)}^{3 / 2} - 30 = 24$
${(3 x - 1)}^{2 / 3} = 4$

Solution

$\begin{array}{rcll} x^{3 / 5} & = & 8 \\ {(x^{3 / 5})}^{5 / 3} & = & 8^{5 / 3} & Raise each side to the power \frac{5}{3} . \\ x & = & 8^{5 / 3} = {(\sqrt[3]{8})}^{5} & {(x^{3 / 5})}^{5 / 3} = 1 \\ x & = & 2^{5} = 32 & \sqrt[3]{8} = 2 \end{array}$

Check: $\begin{array}{rcll} x^{3 / 5} & = & 8 \\ 32^{3 / 5} & \overset{?}{=} & 8 & Replace x with 32 . \\ {(\sqrt[5]{32})}^{3} & \overset{?}{=} & 8 \\ 2^{3} & \overset{?}{=} & 8 ✓ & Yes \end{array}$
$\begin{array}{rcll} 2 {(2 x - 1)}^{3 / 2} - 30 & = & 24 & Original equation \\ 2 {(2 x - 1)}^{3 / 2} & = & 24 + 30 & Add 30 to both sides and simplify . \\ = & 54 & Simplify . \\ {(2 x - 1)}^{3 / 2} & = & \frac{54}{2} = 27 & Divide both sides by 2 . \\ {({(2 x - 1)}^{3 / 2})}^{2 / 3} & = & {(27)}^{2 / 3} & Raise both sides to the power \frac{2}{3} with \\ (odd) m = 3 . \\ 2 x - 1 & = & 9 & {(27)}^{2 / 3} = {(\sqrt[3]{27})}^{2} = 3^{2} = 9 \\ 2 x & = & 10 & Isolate x . \\ x & = & 5 & Solve for x . \end{array}$

Check: $\begin{array}{rcll} 2 {(2 x - 1)}^{3 / 2} - 30 & = & 24 \\ 2 {(2 \cdot 5 - 1)}^{3 / 2} - 30 & \overset{?}{=} & 24 & Replace x with 5 . \\ 2 {(9)}^{3 / 2} - 30 & \overset{?}{=} & 24 & 2 \cdot 5 - 1 = 9 \\ 2 (27) - 30 & \overset{?}{=} & 24 & {(9)}^{3 / 2} = {(\sqrt{9})}^{3} = {(3)}^{3} = 27 \\ 54 - 30 & \overset{?}{=} & 24 ✓ & Yes \end{array}$
$\begin{array}{rcll} {(3 x - 1)}^{2 / 3} & = & 4 & Original equation \\ {[{(3 x - 1)}^{2 / 3}]}^{3 / 2} & = & \pm {(4)}^{3 / 2} & Raise each side to the power \frac{3}{2} . Because the \\ denominator, 2, is even, we use the \pm sign . \\ 3 x - 1 & = & \pm 8 & {(4)}^{3 / 2} = {(\sqrt{4})}^{3} = {(2)}^{3} = 8 \\ 3 x & = & 1 \pm 8 & Add 1 to both sides . \end{array}$

Then

$\begin{array}{c} or \\ \begin{array}{rclrcl} 3 x & = & 1 + 8 & 3 x & = & 1 - 8 \\ 3 x & = & 9 & 3 x & = & 7 \\ x & = & 3 & x & = & - \frac{7}{3} \end{array} \end{array}$

Verify that both $x = 3$ and $x = - \frac{7}{3}$ satisfy the original equation. The solution set is ${- \frac{7}{3}, 3} .$

Practice Problem 8

Solve:
1. $3 {(x - 2)}^{3 / 5} + 4 = 7$
2. ${(2 x + 1)}^{4 / 3} - 7 = 9$

Equations That Are Quadratic in Form

5 Solve equations that are quadratic in form.

An equation in a variable x is quadratic in form if it can be written as

$a u^{2} + b u + c = 0 (a \neq 0),$

where u is an expression in the variable x We solve the equation $a u^{2} + b u + c = 0$ for u Then the solutions of the original equation can be obtained by replacing u with the expression in x that u represents.

Example 9 Solving an Equation Quadratic in Form

Solve: $x^{2 / 3} - 5 x^{1 / 3} + 6 = 0$

Solution

The given equation is not a quadratic equation. However, if we let $u = x^{1 / 3},$ then $u^{2} = {(x^{1 / 3})}^{2} = x^{2 / 3} .$ The original equation becomes a quadratic equation if we replace $x^{1 / 3}$ with u.

$\begin{array}{rcll} x^{2 / 3} - 5 x^{1 / 3} + 6 & = & 0 & Original equation \\ u^{2} - 5 u + 6 & = & 0 & Replace x^{1 / 3} with u . \\ (u - 2) (u - 3) & = & 0 & Factor . \\ u - 2 = 0 or u - 3 & = & 0 & Zero-product property \\ u = 2 or u & = & 3 & Solve for u . \end{array}$

Because $u = x^{1 / 3},$ we find x from the equations $u = 2$ and $u = 3 .$

$\begin{array}{rcllrcl} x^{1 / 3} & = & 2 & or & x^{1 / 3} & = & 3 & Replace u with x^{1 / 3} . \\ x & = & 2^{3} & or & x & = & 3^{3} & Here \frac{m}{n} = \frac{1}{3}, so m = 1 is odd . \\ x & = & 8 & or & x & = & 27 are & two apparent solutions . \end{array}$

Check: You should check that both $x = 8$ and $x = 27$ satisfy the equation. The solution set is {8, 27}.

Practice Problem 9

Solve: $x^{2 / 3} - 7 x^{1 / 3} + 6 = 0$

Warning

When u replaces an expression in the variable x in an equation that is quadratic in form, you are not finished once you have found values for u You then must replace u with the expression in x and solve for x.

Example 10 Solving an Equation That Is Quadratic in Form

Solve: ${(x + \frac{1}{x})}^{2} - 6 (x + \frac{1}{x}) + 8 = 0$

Solution

We let $u = x + \frac{1}{x} .$ Then $u^{2} = {(x + \frac{1}{x})}^{2} .$ With this substitution, the original equation becomes a quadratic equation in u.

$\begin{array}{l} \begin{array}{rcll} {(x + \frac{1}{x})}^{2} - 6 (x + \frac{1}{x}) + 8 & = & 0 & Original equation \\ u^{2} - 6 u + 8 & = & 0 & Replace x + \frac{1}{x} with u . \\ (u - 2) (u - 4) & = & 0 & Factor . \end{array} \\ \begin{array}{rcrl} u - 2 = 0 & or & u - 4 = 0 & Zero-product property \\ u = 2 & or & u = 4 & Solve for u . \end{array} \end{array}$

Replacing u with $x + \frac{1}{x}$ in these equations, we have

$x + \frac{1}{x} = 2 or x + \frac{1}{x} = 4 .$

Next, we solve each of these equations for x.

$\begin{array}{rcll} x + \frac{1}{x} & = & 2 \\ x^{2} + 1 & = & 2 x & Multiply both sides by x (the LCD) . \\ x^{2} - 2 x + 1 & = & 0 & Subtract 2 x from both sides; simplify . \\ {(x - 1)}^{2} & = & 0 & Factor the perfect-square trinomial . \\ x - 1 & = & 0 & Square root property \\ x & = & 1 & Solve for x . \end{array}$

Thus, $x = 1$ is a possible solution of the original equation. Now we solve:

$\begin{array}{rcll} x + \frac{1}{x} & = & 4 \\ x^{2} + 1 & = & 4 x & Multiply both sides by x (LCD) . \\ x^{2} - 4 x + 1 & = & 0 & Subtract 4 x from both sides . \\ x & = & \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (1)}}{2 (1)} & \begin{array}{l} Quadratic formula: \\ a = 1, b = - 4, c = 1 \end{array} \\ x & = & \frac{4 \pm \sqrt{12}}{2} & Simplify . \\ x & = & \frac{4 \pm 2 \sqrt{3}}{2} & \sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \sqrt{3} = 2 \sqrt{3} \\ x & = & \frac{2 (2 \pm \sqrt{3})}{2} & Factor the numerator . \\ x & = & 2 \pm \sqrt{3} & Remove the common factor . \end{array}$

Thus, $2 + \sqrt{3}$ and $2 - \sqrt{3}$ are also possible solutions of the original equation. Consequently, the possible solutions of the original equation are $1, 2 + \sqrt{3},$ and $2 - \sqrt{3} .$

You should verify that 1, $2 + \sqrt{3},$ and $2 - \sqrt{3}$ are solutions of the original equation. The solution set is ${1, 2 - \sqrt{3}, 2 + \sqrt{3}} .$

Practice Problem 10

Solve: ${(1 + \frac{1}{x})}^{2} - 6 (1 + \frac{1}{x}) + 8 = 0$

The next example illustrates the paradoxical effect of time dilation (page 130) when one travels at speeds that are close to the speed of light.

Example 11 Investigating Space Travel (Your Older Sister Is Younger Than You Are)

Your sister is five years older than you are. She decides that she has had enough of Earth and needs a vacation. She takes a trip to the Omega-One star system. Her trip to Omega-One and back in a spacecraft traveling at an average speed v took 15 years according to the clock and calendar on the spacecraft. But upon landing back on Earth, she discovers that her voyage took 25 years according to the time on Earth. This means that although you were five years younger than your sister before her vacation, you are five years older than her after her vacation! Use the time dilation equation $t_{0} = t \sqrt{1 - \frac{v^{2}}{c^{2}}}$ from the introduction to this section to calculate the speed of the spacecraft.

Solution

Substitute $t_{0} = 15$ (moving-frame time) and $t = 25$ (fixed-frame time) to obtain

$\begin{array}{rcll} 15 & = & 25 \sqrt{1 - \frac{v^{2}}{c^{2}}} \\ \frac{3}{5} & = & \sqrt{1 - \frac{v^{2}}{c^{2}}} & Divide both sides by 25; \frac{15}{25} = \frac{3}{5} . \\ \frac{9}{25} & = & 1 - \frac{v^{2}}{c^{2}} & Square both sides . \\ \frac{v^{2}}{c^{2}} & = & 1 - \frac{9}{25} & \begin{array}{l} Add \frac{v^{2}}{c^{2}} - \frac{9}{25} to both sides . \end{array} \\ {(\frac{v}{c})}^{2} & = & \frac{16}{25} & Simplify . \\ \frac{v}{c} & = & \pm \frac{4}{5} & Square root property \\ \frac{v}{c} & = & \frac{4}{5} & Reject the negative value . \\ v & = & \frac{4}{5} c = 0.8 c & Multiply both sides by c . \end{array}$

Thus, the spacecraft must have been traveling at 80% of the speed of light $(0.8 c)$ when your sister found the “trip of youth.”

Practice Problem 11

Suppose your sister’s trip took 20 years according to the clock and calendar on the spacecraft, but 25 years judging by time on Earth. Find the speed of the spacecraft.

Section 1.5 Exercises

Concepts and Vocabulary

If an apparent solution does not satisfy the equation, it is called a(n) solution.
When solving a rational equation, we multiply both sides by the .
If $x^{3 / 4} = 8,$ then $x = \underline{}$ .
If $x^{4 / 3} = 16,$ then $x = \underline{}$ or $x = \underline{}$ .
True or False. Rational equations always have extraneous solutions.
True or False. If $x^{2 / 3} = k,$ then $x = \pm \sqrt{k^{3}} .$
True or False. The equations $x = 3$ and $x^{2} = 9$ have identical solutions.
True or False. The first step in solving equations involving radicals is to isolate one radical on one side of the equation.

Building Skills

In Exercises 9–18, find the real solutions of each equation by factoring.

$x^{3} = 2 x^{2}$
$3 x^{4} - 27 x^{2} = 0$
${(\sqrt{x})}^{3} = \sqrt{x}$
${(\sqrt{x})}^{5} = 16 \sqrt{x}$
$x^{3} + x = 0$
$x^{3} - 1 = 0$
$x^{4} - x^{3} = x^{2} - x$
$x^{3} - 36 x = 16 (x - 6)$
$x^{4} = 27 x$
$3 x^{4} = 24 x$

In Exercises 19–28, solve each equation by multiplying both sides by the LCD.

$\frac{x + 1}{3 x - 2} = \frac{5 x - 4}{3 x + 2}$
$\frac{x}{2 x + 1} = \frac{3 x + 2}{4 x + 3}$
$\frac{1}{x} + \frac{2}{x + 1} = 1$
$\frac{x}{x + 1} + \frac{x + 1}{x + 2} = 1$
$\frac{6 x - 7}{x} - \frac{1}{x^{2}} = 5$
$\frac{1}{x} + \frac{2}{x + 1} + \frac{3}{x + 2} = 0$
$\frac{1}{x} + \frac{1}{x - 3} = \frac{7}{3 x - 5}$
$\frac{x + 3}{x - 1} + \frac{x + 4}{x + 1} = \frac{8 x + 5}{x^{2} - 1}$
$\frac{5}{x + 1} - \frac{4}{2 x + 2} + \frac{2}{2 x - 1} = \frac{13}{18}$
$\frac{6}{2 x - 2} - \frac{1}{x + 1} = \frac{2}{2 x + 2} + 1$

In Exercises 29–36, solve each equation. Check your answers.

$\frac{x}{x - 5} - \frac{5}{x + 5} = \frac{10 x}{x^{2} - 25}$
$\frac{x}{x - 4} - \frac{4}{x + 4} = \frac{8 x}{x^{2} - 16}$
$\frac{x}{x - 3} + \frac{3}{x + 3} = \frac{6 x}{x^{2} - 9}$
$\frac{x}{x + 7} + \frac{7}{x - 7} = \frac{14 x}{x^{2} - 49}$
$\frac{1}{x - 1} + \frac{x}{x + 3} = \frac{4}{x^{2} + 2 x - 3}$
$\frac{x}{x - 2} + \frac{2}{x + 3} = - \frac{10}{x^{2} + x - 6}$
$\frac{2 x}{x + 3} - \frac{x}{x - 1} = \frac{14}{x^{2} + 2 x - 3}$
$\frac{2 (x + 1)}{x - 2} - \frac{x}{x + 1} = \frac{9}{x^{2} - x - 2}$

In Exercises 37–58, solve each equation. Check your answers.

$\sqrt[3]{3 x - 1} = 2$
$\sqrt[3]{2 x + 3} = 3$
$\sqrt{x - 1} = - 2$
$\sqrt{3 x + 4} = - 1$
$x + \sqrt{x + 6} = 0$
$x - \sqrt{6 x + 7} = 0$
$\sqrt{y + 6} = y$
$r + 11 = 6 \sqrt{r + 3}$
$\sqrt{6 y - 11} = 2 y - 7$
$\sqrt{3 y + 1} = y - 1$
$t - \sqrt{3 t + 6} = - 2$
$\sqrt{5 x^{2} - 10 x + 9} = 2 x - 1$
$\sqrt{x - 3} = \sqrt{2 x - 5} - 1$
$x + \sqrt{x + 1} = 5$
$\sqrt{2 y + 9} = 2 + \sqrt{y + 1}$
$\sqrt{m} - 1 = \sqrt{m - 5}$
$\sqrt{7 z + 1} - \sqrt{5 z + 4} = 1$
$\sqrt{3 q + 1} - \sqrt{q - 1} = 2$
$\sqrt{2 x + 5} + \sqrt{x + 6} = 3$
$\sqrt{5 x - 9} - \sqrt{x + 4} = 1$
$\sqrt{2 x - 5} - \sqrt{x - 3} = 1$
$\sqrt{3 x + 5} + \sqrt{6 x + 3} = 3$

In Exercises 59–62, solve each equation. Check your answers.

${(x - 4)}^{3 / 2} = 27$
${(x + 7)}^{3 / 2} = 64$
${(5 x - 3)}^{2 / 3} - 5 = 4$
${(2 x - 6)}^{2 / 3} + 9 = 13$

In Exercises 63–86, solve each equation by using an appropriate substitution. Check your answers.

$x - 5 \sqrt{x} + 6 = 0$
$x - 3 \sqrt{x} + 2 = 0$
$2 y - 15 \sqrt{y} = - 7$
$y + 44 = 15 \sqrt{y}$
$x^{- 2} - x^{- 1} - 42 = 0$
$x^{1 / 2} + 3 - 4 x^{- 1 / 2} = 0$
$x^{2 / 3} - 6 x^{1 / 3} + 8 = 0$
$x^{2 / 5} + x^{1 / 5} - 2 = 0$
$2 x^{1 / 2} + 3 x^{1 / 4} - 2 = 0$
$2 x^{1 / 2} - x^{1 / 4} - 1 = 0$
$x^{4} - 13 x^{2} + 36 = 0$
$x^{4} - 7 x^{2} + 12 = 0$
$2 t^{4} + t^{2} - 1 = 0$
$81 y^{4} + 1 = 18 y^{2}$
$p^{2} - 3 + 4 \sqrt{p^{2} - 3} - 5 = 0$
$x^{2} - 3 - 4 \sqrt{x^{2} - 3} - 12 = 0$
${(3 t + 1)}^{2} - 3 (3 t + 1) + 2 = 0$
${(7 z + 5)}^{2} + 2 (7 z + 5) - 15 = 0$
${(\sqrt{y} + 5)}^{2} - 9 (\sqrt{y} + 5) + 20 = 0$
${(2 \sqrt{t} + 1)}^{2} - 2 (2 \sqrt{t} + 1) - 3 = 0$
${(x^{2} - 4)}^{2} - 3 (x^{2} - 4) - 4 = 0$
${(x^{2} - 1)}^{2} - 11 (x^{2} - 1) + 24 = 0$
${(x^{2} - 3 x)}^{2} - 2 (x^{2} - 3 x) - 8 = 0$
${(x^{2} - 4 x)}^{2} + 7 (x^{2} - 4 x) + 12 = 0$

Applying the Concepts

Work efficiency. Working together, Todrick and his brother can finish laying concrete blocks in 2 days. If each works alone, Todrick’s brother takes 3 more days than it would take Todrick to do the job. How long would it take each of them working alone to finish the job?
Work efficiency. Working together, Celeste and Jacob shovel the snow from their driveway in 6 hours. It takes Jacob five hours more than Celeste to do the job alone. How long would it take each of them working alone to finish the job?
Pressure in a hose. The pressure in a sprinkler hose varies with time according to the equation $p = 63 + \sqrt{t + 40}$ psi. At what time will the pressure be 70 psi?
Geometric dimensions. Find the dimensions of a rectangle with an area of 12 sq ft and a diagonal of 5 ft.
Fractions. The numerator of a fraction is two less than the denominator. The sum of the fraction and its reciprocal is $\frac{25}{12} .$ Find the numerator and denominator assuming that each is a positive integer.
Fractions. The numerator of a fraction is five less than the denominator. The sum of the fraction and six times its reciprocal is $\frac{25}{2} .$ Find the numerator and denominator, assuming that each is an integer.
Investment. Latasha bought some stock for $1800. If the price of each share of the stock had been $18 less, she could have bought five more shares for the same $1800.
1. How many shares of the stock did she buy?
2. How much did she pay for each share?
Bus charter. A civic club charters a bus for one day at a cost of $575. When two more people join the group, each person’s cost decreases by $2. How many people were in the original group?
Depth of a well. A stone is dropped into a well, and the time it takes to hear the splash is 4 seconds. Find the depth of the well (to the nearest foot). Assume that the speed of sound is 1100 feet per second and that $d = 16 t^{2}$ is the distance d an object falls in t seconds.
Estimating the horizon. You are in a hot air balloon 2 miles above the ocean. (See the accompanying diagram.) As far as you can see, there is only water. Given that the radius of the earth is 3960 miles, how far away is the horizon?
Rate of current. A motorboat is capable of traveling at a speed of 10 miles per hour in still water. On a particular day, the boat took 30 minutes longer to travel a distance of 12 miles upstream than it took to travel the same distance downstream. What was the rate of the current in the stream on that day?
Train speed. A freight train requires $2 \frac{1}{2}$ hours longer to make a 300-mile journey than an express train does. If the express train averages 20 miles per hour faster than the freight train, how long does it take the express train to make the trip?
Washing the family car. A couple washed the family car in 24 minutes. Previously, when they each had washed the car alone, it took the husband 20 minutes longer to wash the car than it took the wife. How long did it take the wife to wash the car?
Filling a swimming pool. A small swimming pool can be filled by two hoses together in 1 hour and 12 minutes. The larger hose alone fills the pool in 1 hour less than the smaller hose. How long does it take the smaller hose to fill the pool?
The area of a rectangular plot. Together the diagonal and the longer side of a rectangular plot are three times the length of the shorter side. If the longer side exceeds the shorter side by 100 feet, what is the area of the plot?
Emergency at sea. You are at sea in your motorboat at a point A located 40 km from a point B that lies 130 kilometers from a hospital on a long, straight road. (See the accompanying figure.) While at point A, your companion feels sick and has to be taken to the hospital. You call the hospital on your cell phone to send an ambulance. Your motorboat can travel at 40 kilometers per hour, and the ambulance can travel at 80 kilometers per hour. How far from point B will you be when you and the ambulance arrive simultaneously at point C?
Planning road construction. Two towns A and B are located 8 and 5 miles, respectively, from a long, straight highway. The points C and D on the highway closest to the two towns are 18 miles apart. (Remember that the shortest distance from a point to a line is the length of the perpendicular segment from the point to the line.) Where should point E be located on the highway so that the sum of the lengths of the roads from A to E and E to B is 23 miles? (See the accompanying figure.)

[Hint: $205 x^{2} - 4392 x + 18972 = (x - 6) (205 x - 3162) .$ ]
Department store purchases. A department store buyer purchased some shirts for $540. Then she sold all but eight of them. On each shirt sold, a profit of $6 (over the purchase price) was made. The shirts sold brought in revenue of $780.
1. How many shirts were bought initially?
2. What was the purchase price of each shirt?
3. What was the selling price of each shirt?

Beyond the Basics

In Exercises 105–114, solve each equation by any method.

${(\frac{x^{2} - 3}{x})}^{2} - 6 (\frac{x^{2} - 3}{x}) + 8 = 0$
${(\frac{x + 2}{3 x + 1})}^{2} - 5 (\frac{x + 2}{3 x + 1}) + 6 = 0$
$\sqrt{\sqrt{3 x - 2} + 2 \sqrt{4 x + 1}} = 2 \sqrt{2}$
$\frac{5}{x - 1} - \frac{3 x}{x + 1} = \frac{9}{x^{2} - 1}$
$\frac{x + 2}{x} + \frac{2 x + 6}{x^{2} + 4 x} = - \frac{1}{x + 4}$
$(x^{2} + \frac{1}{x^{2}}) - 7 (x - \frac{1}{x}) + 8 = 0$

[Hint: $u = x - \frac{1}{x}, u^{2} + 2 = x^{2} + \frac{1}{x^{2}} .$ ]
${(\frac{t}{t + 2})}^{2} + \frac{2 t}{t + 2} - 15 = 0$
$2 x^{1 / 3} + 2 x^{- 1 / 3} - 5 = 0$
$x^{4 / 3} - 4 x^{2 / 3} + 3 = 0$
$8 \sqrt{\frac{x}{x + 3}} - \sqrt{\frac{x + 3}{x}} = 2$

In Exercises 115 and 116, solve for x in terms of the other variables. (All letters denote positive real numbers.)

$\frac{x + b}{x - b} = \frac{x - 5 b}{2 x - 5 b}$
$3 a - \sqrt{a x} = \sqrt{a (3 x + a)}$

Critical Thinking/Discussion/Writing

In Exercises 117–119, find the value of x.

$x = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}$

[Hint: $x = \sqrt{1 + x} .$ ]
$x = \sqrt{20 + \sqrt{20 + \sqrt{20 + \dots}}}$
$x = \sqrt{n + \sqrt{n + \sqrt{n + \dots}}},$ where n is a positive integer.
Assume that $x > 1 .$ Solve for x.

$\sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}} = 4$

[Hint: $\begin{array}{rcl} x + 2 \sqrt{x - 1} & = & (x - 1) + 1 + 2 \sqrt{x - 1} \\ = & {(\sqrt{x - 1} + 1)}^{2} .] \end{array}$

Getting Ready for the Next Section

In Exercises 121–127, use inequality symbols to write the statement symbolically.

5 is greater than $- 1 .$
$- 4$ is less than $- 1 .$
7 is greater than or equal to 7.
2x is less than $2 x + 5$ .
3 is greater than $3 x + 4 .$
$x - 1$ is less than or equal to 2.
x is nonnegative.

In Exercises 128–132, solve each equation.

$12 x - 4 = 28 - 4 x$
$2 (x - 3) + 17 = 4 x + 1$
$2 x + 3 (x - 4) = 7 x + 10$
$2 x - 3 - (3 x - 1) = 6$
$\frac{- 2 x}{3} = x + \frac{x}{2}$

In Exercises 133–136, perform the indicated operations, factoring all denominators.

$\frac{5}{2 x} + 3$
$6 x + \frac{5}{x - 3}$
$\frac{3}{x + 1} - \frac{x - 2}{x + 3}$
$\frac{x}{x^{2} + 6 x + 9} - \frac{x - 3}{x^{2} + 5 x + 6}$

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Table of Contents for Section 1.5 Solving Other Types of Equations

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Table of Contents for
Section 1.5 Solving Other Types of Equations