please answer a minimum of 3 OF 6 question pairs. you must answer one question out of each pair with a minimum of one paragraph.
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Ch1lecture.pptxCopyright © Cengage Learning. All rights reserved.
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Copyright © Cengage Learning. All rights reserved.
Linear and Absolute Value
Equations
SECTION 1.1
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Linear Equations
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Linear Equations
An equation is a statement about the equality of two expressions.
If either of the expressions contains a variable, the equation may be a true statement for some values of the variable and a false statement for other values.
For example, the equation 2x + 1 = 7 is a true statement for x = 3, but it is false for any number except 3.
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Linear Equations
The number 3 is said to satisfy the equation 2x + 1 = 7 because substituting 3 for x produces 2(3) + 1 = 7, which is a true statement.
To solve an equation means to find all values of the variable that satisfy the equation.
The values that satisfy an equation are called solutions or roots of the equation.
For instance, 2 is a solution of x + 3 = 5.
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Linear Equations
Equivalent equations are equations that have exactly the same solution or solutions.
The process of solving an equation is often accomplished by producing a sequence of equivalent equations until we arrive at an equation or equations of the form
Variable = Constant
To produce these equivalent equations, apply the properties of real numbers and the following two properties of equality.
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Linear Equations
Addition and Subtraction Property of Equality
Adding the same expression to each side of an equation or subtracting the same expression from each side of an equation produces an equivalent equation.
Example
Begin with the equation 2x – 7 = 11.
Replacing x with 9 shows that 9 is a solution of the equation. Now add 7 to each side of the equation.
The resulting equation is 2x = 18, and the solution of the new equation is still 9.
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Linear Equations
Multiplication and Division Property of Equality
Multiplying or dividing each side of an equation by the same nonzero expression produces an equivalent equation.
Example
Begin with the equation x = 8.
Replacing x with 12 shows that 12 is a solution of the equation. Now multiply each side of the equation by .
The resulting equation is x = 12, and the solution of the new equation is still 12.
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Linear Equations
Many applications can be modeled by linear equations in one variable.
Definition of a Linear Equation
A linear equation, or first-degree equation, in the single variable x is an equation that can be written in the form
ax + b = 0
where a and b are real numbers, with a 0.
Linear equations are solved by applying the properties of real numbers and the properties of equality.
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Example 1 – Solve a Linear Equation in One Variable
Solve: 3x – 5 = 7x – 11
Solution:
3x – 5 = 7x – 11
3x – 7x – 5 = 7x – 7x – 11
–4x – 5 = –11
–4x – 5 + 5 = –11 + 5
–4x = –6
Subtract 7x from each side of the equation.
Add 5 to each side of the equation.
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Example 1 – Solution
The solution is .
Divide each side of the equation by –4.
The equation is now in the form
Variable = Constant.
cont’d
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Linear Equations
When an equation contains parentheses, use the distributive property to remove the parentheses.
If an equation involves fractions, it is helpful to multiply each side of the equation by the least common denominator (LCD) of all denominators to produce an equivalent equation that does not contain fractions.
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Contradictions, Conditional Equations,
and Identities
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Contradictions, Conditional Equations, and Identities
An equation that has no solutions is called a contradiction.
The equation x = x + 1 is a contradiction. No number is equal to itself increased by 1.
An equation that is true for some values of the variable but not true for other values of the variable is called a conditional equation.
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Contradictions, Conditional Equations, and Identities
For example, x + 2 = 8 is a conditional equation because it is true for x = 6 and false for any number not equal to 6.
An identity is an equation that is true for all values of the variable for which all terms of the equation are defined.
Examples of identities include the equations
x + x = 2x and 4(x + 3) – 1 = 4x + 11.
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Example 4 – Classify Equations
Classify each equation as a contradiction, a conditional equation, or an identity.
a. x + 1 = x + 4
b. 4x + 3 = x – 9
c. 5(3x – 2) – 7(x – 4) = 8x + 18
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Example 4 – Solution
a. Subtract x from both sides of x + 1 = x + 4 to produce the equivalent equation 1 = 4.
Because 1 = 4 is a false statement, the original equation x + 1 = x + 4 has no solutions.
It is a contradiction.
b. Solve using the procedures that produce equivalent equations. 4x + 3 = x – 9
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Example 4 – Solution
3x + 3 = –9
3x = –12
x = –4
Check to confirm that –4 is a solution.
The equation 4x + 3 = x – 9 is true for x = –4, but it is not true for any other values of x.
Thus 4x + 3 = x – 9 is a conditional equation.
Subtract x from each side.
Subtract 3 from each side.
Divide each side by 3.
cont’d
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Example 4 – Solution
c. Simplify the left side of the equation to show that it is identical to the right side.
5(3x – 2) – 7(x – 4) = 8x + 18
15x – 10 – 7x + 28 = 8x + 18
8x + 18 = 8x + 18
The original equation 5(3x – 2) – 7(x – 4) = 8x + 18 is true for all real numbers x.
The equation is an identity.
cont’d
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Absolute Value Equations
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Absolute Value Equations
The absolute value of a real number x is the distance between the number x and the number 0 on the real number line.
Thus the solutions of | x | = 3 are all real numbers that are
3 units from 0.
Therefore, the solutions of | x | = 3 are x = 3 or x = –3. See Figure 1.1.
| x | = 3
Figure 1.1
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Absolute Value Equations
The following property is used to solve absolute value equations.
A Property of Absolute Value Equations
For any variable expression E and any nonnegative real number k,
| E | = k if and only if E = k or E = –k
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Absolute Value Equations
Example
If | x | = 5, then x = 5 or x = –5.
If | x | = , then x = or x = – .
If | x | = 0, then x = 0.
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Example 5 – Solve an Absolute Value Equation
Solve: | 2x – 5 | = 21
Solution:
| 2x – 5 | = 21 implies 2x – 5 = 21 or 2x – 5 = –21.
Solving each of these linear equations produces
2x – 5 = 21 or 2x – 5 = –21
2x = 26 2x = –16
x = 13 x = –8
The solutions are –8 and 13.
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Applications of Linear Equations
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Example 6 – Movie Theater Ticket Prices
Movie theater ticket prices have been increasing steadily in recent years (see Table 1.1).
Table 1.1
Average U.S. Movie Theater Ticket Price
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Example 6 – Movie Theater Ticket Prices
An equation that models the average U.S. movie theater
ticket price p, in dollars, is given by
p = 0.293t + 6.590
where t is the number of years after 2006. (This means that t = 0 corresponds to 2006.) Use this equation to predict the year in which the average U.S. movie theater ticket price will reach $9.00.
cont’d
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Example 6 – Solution
p = 0.293t + 6.590
9.00 = 0.293t + 6.590
2.41 = 0.293t
t 8.2
Our equation predicts that the average U.S. movie theater ticket price will reach $9.00 about 8.2 years after 2006, which is 2014.
Substitute 9.00 for p.
Solve for t.
cont’d
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Formulas and Applications
SECTION 1.2
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Formulas
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Formulas
A formula is an equation that expresses known relationships between two or more variables.
Table 1.2 lists several formulas from geometry that are used in this text.
The variable P represents perimeter, C represents circumference of a circle, A represents area, S represents surface area of an enclosed solid, and V represents volume.
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Formulas
Table 1.2
Formulas from Geometry
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Formulas
It is often necessary to solve a formula for a specified variable.
Begin the process by isolating all terms that contain the specified variable on one side of the equation and all terms that do not contain the specified variable on the other side.
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Example 1 – Solve a Formula for a Specified Variable
a. Solve 2l + 2w = P for l.
b. Solve S = 2(wh + lw + hl) for h.
Solution:
a. 2l + 2w = P
2l = P – 2w
Subtract 2w from each side to isolate the 2l term.
Divide each side by 2.
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Example 1 – Solution
b. S = 2(wh + lw + hl)
S = 2wh + 2lw + 2hl
S – 2lw = 2wh + 2hl
S – 2lw = 2h(w + l)
cont’d
Isolate the terms that involve the variable h on the right side.
Factor 2h from the right side.
Divide each side by 2(w + l ).
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Applications
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Applications
Linear equations emerge in a variety of application problems. In solving such problems, it generally helps to apply specific techniques in a series of small steps.
The following general strategies should prove helpful in the remaining portion of this section.
Strategies for Solving Application Problems
1. Read the problem carefully. If necessary, reread the problem several times.
2. When appropriate, draw a sketch and label parts of the drawing with the specific information given in the problem.
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Applications
3. Determine the unknown quantities, and label them with variables. Write down any equation that relates the variables.
4. Use the information from step 3, along with a known formula or some additional information given in the problem, to write an equation.
5. Solve the equation obtained in step 4, and check to see whether the results satisfy all the conditions of the original problem.
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Example 3 – Dimensions of a Painting
A dog agility course must be contained inside a rectangle where the ratio of the length to the width is 1.2. If there are 440 feet of fencing available to enclose the course, what are the dimensions of the rectangle?
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Example 3 – Solution
1. Read the problem carefully.
2. Draw a rectangle and label it. We have used w for its width and l for its length. See Figure 1.3.
3. The problem states that the ratio of length to width is 1.2. Write this as an equation.
or, multiplying each side of the equation by w, l = 1.2w.
Figure 1.3
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Example 3 – Solution
The problem also states that the perimeter is 440 feet. Write this as an equation. The formula for perimeter is P = 2l + 2w.
4. Combine the information from the preceding steps to write an equation.
P = 2l + 2w
440 = 2(1.2w) + 2w
cont’d
P = 440, l = 1.2w
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Example 3 – Solution
5. Solve for w.
440 = 2(1.2w) + 2w
440 = 2.4w + 2w
440 = 4.4w
100 = w
The width is 100 feet. The length is l = 1.2(100) = 120.
The width of the rectangle is 100 feet, and the length is 120 feet.
cont’d
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Applications
Similar triangles are ones for which the measures of corresponding angles are equal. The triangles below are similar.
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Applications
An important relationship among the sides of similar triangles is that the ratios of corresponding sides are equal.
Thus, for the triangles,
This fact is used in many applications.
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Applications
Many business applications can be solved by using the equation
Profit = revenue – cost
Simple interest problems can be solved by using the formula I = Prt, where I is the interest, P is the principal, r is the simple interest rate per period, and t is the number of periods.
Many uniform motion problems can be solved by using the formula d = rt, where d is the distance traveled, r is the rate of speed, and t is the time.
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Applications
Percent mixture problems involve combining solutions or alloys that have different concentrations of a common substance.
Percent mixture problems can be solved by using the formula pA = Q, where p is the percent of concentration (in decimal form), A is the amount of the solution or alloy, and Q is the quantity of a substance in the solution or alloy.
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Applications
For example, in 4 liters of a 25% acid solution, p is the percent of acid (0.25 as a decimal), A is the amount of solution (4 liters), and Q is the amount of acid in the solution, which equals (0.25)(4) liters = 1 liter.
Value mixture problems involve combining two or more ingredients that have different prices into a single blend.
The solution of a value mixture problem is based on the equation V = CA, where V is the value of the ingredient, C is the unit cost of the ingredient, and A is the amount of the ingredient.
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Applications
For instance, if the cost C of tea is $4.30 per pound, then 5 pounds (the amount A) of tea has a value V = (4.30)(5) = 21.50, or $21.50.
The solution of a value mixture problem is based on the sum of the values of all ingredients taken separately equaling the value of the mixture.
To solve a work problem, use the equation
Rate of work Time worked = Part of task completed
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Applications
For example, if a painter can paint a wall in 15 minutes,
then the painter can paint of the wall in 1 minute.
The painter’s rate of work is of the wall each minute.
In general, if a task can be completed in x minutes, then
the rate of work is of the task each minute.
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Quadratic Equations
SECTION 1.3
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Solving Quadratic Equations by Factoring
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Solving Quadratic Equations by Factoring
In this section we will learn to solve a type of equation that is referred to as a quadratic equation.
Definition of a Quadratic Equation
A quadratic equation in x is an equation that can be written in the standard quadratic form
ax2 + bx + c = 0
where a, b and c are real numbers and a 0.
Several methods can be used to solve a quadratic equation.
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Solving Quadratic Equations by Factoring
For instance, if you can factor ax2 + bx + c into linear factors, then ax2 + bx + c = 0 can be solved by applying the following property.
The Zero Product Principle
If A and B are algebraic expressions such that AB = 0, then A = 0 or B = 0.
The zero product principle states that if the product of two factors is 0, then at least one of the factors must be 0. In Example 1, the zero product principle is used to solve a quadratic equation.
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Example 1 – Solve by Factoring
Solve each quadratic equation by factoring.
a. x2 + 2x – 15 = 0 b. 2x2 – 5x = 12
Solution:
a. x2 + 2x – 15 = 0
(x – 3)(x + 5) = 0
x – 3 = 0 or x + 5 = 0
x = 3 x = –5
A check shows that 3 and –5 are both solutions of x2 + 2x – 15 = 0.
Factor.
Set each factor equal to 0.
Solve each linear equation.
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Example 1 – Solution
b. 2x2 – 5x = 12
2x2 – 5x – 12 = 0
(x – 4)(2x + 3) = 0
x – 4 = 0 or 2x + 3 = 0
x = 4 2x = –3
x =
A check shows that 4 and are both solutions of
2x2 – 5x = 12.
cont’d
Write in standard quadratic form.
Factor.
Set each factor equal to 0.
Solve each linear equation.
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Solving Quadratic Equations by Factoring
Some quadratic equations have a solution that is called a double root. For instance, consider x2 – 8x + 16 = 0.
Solving this equation by factoring, we have
x2 – 8x + 16 = 0
(x – 4)(x – 4) = 0
x – 4 = 0 or x – 4 = 0
x = 4 x = 4
Factor.
Set each factor equal to 0.
Solve each linear equation.
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Solving Quadratic Equations by Factoring
Some quadratic equations have a solution that is called a double root. For instance, consider x2 – 8x + 16 = 0.
Solving this equation by factoring, we have
x2 – 8x + 16 = 0
(x – 4)(x – 4) = 0
x – 4 = 0 or x – 4 = 0
x = 4 x = 4
Factor.
Set each factor equal to 0.
Solve each linear equation.
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Solving Quadratic Equations by Taking Square Roots
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Solving Quadratic Equations by Taking Square Roots
Recall that This principle can be used to solve some quadratic equations by taking the square root of each side of the equation.
In the following example, we use this idea to solve x2 = 25.
x2 = 25
=
| x | = 5
x = –5 or x = 5
The solutions are –5 and 5.
Take the square root of each side.
Use the fact that = | x | and = 5.
Solve the absolute value equation.
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Solving Quadratic Equations by Taking Square Roots
We will refer to the preceding method of solving a quadratic equation as the square root procedure.
The Square Root Procedure
If x2 = c, then x = or x = , which can also be written as x = .
Example
If x2 = 9, then x = = 3 or x = = –3.
This can be written as x = 3.
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Solving Quadratic Equations by Taking Square Roots
If x2 = 7, then x = or x = .
This can be written as x = .
If x2 = –4, then x = = 2i or x = = –2i.
This can be written as x = 2i.
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Example 2 – Solve by Using the Square Root Procedure
Use the square root procedure to solve each equation.
a. 3x2 + 12 = 0 b. (x + 1)2 = 48
Solution:
a. 3x2 + 12 = 0
3x2 = –12
x2 = –4
x =
Solve for x2.
Take the square root of each side of the equation and insert a plus-or-minus sign in front of the radical.
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Example 2 – Solution
x = –2i or x = 2i
The solutions are –2i and 2i.
b. (x + 1)2 = 48
cont’d
Take the square root of each side of the equation and insert a plus-or-minus sign in front of the radical.
Simplify.
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Example 2 – Solution
The solutions are and .
cont’d
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Solving Quadratic Equations by Completing the Square
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Solving Quadratic Equations by Completing the Square
Consider two binomial squares and their perfect-square trinomial products.
In each of the preceding perfect-square trinomials, the coefficient of x2 is 1 and the constant term is the square of half the coefficient of the x term.
x2 + 10x + 25,
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Solving Quadratic Equations by Completing the Square
x2 – 6x + 9,
Adding to a binomial of the form x2 + bx the constant term that makes the binomial a perfect-square trinomial is called completing the square.
For example, to complete the square of x2 + 8x, add
to produce the perfect-square trinomial x2 + 8x + 16.
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Solving Quadratic Equations by Completing the Square
Completing the square is a powerful procedure that can be used to solve any quadratic equation.
For instance, to solve x2 – 6x + 13 = 0, first isolate the variable terms on one side of the equation and the constant term on the other side.
x2 – 6x = –13
x2 – 6x + 9 = –13 + 9
Subtract 13 from each side of the equation.
Complete the square by adding
to each side of the
equation.
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Solving Quadratic Equations by Completing the Square
(x – 3)2 = –4
x – 3 =
x – 3 = 2i
x = 3 2i
The solutions of x2 – 6x + 13 = 0 are 3 – 2i and 3 + 2i.
You can check these solutions by substituting each solution into the original equation.
Factor and solve by the square root procedure.
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Solving Quadratic Equations by Completing the Square
For instance, the following check shows that 3 – 2i does satisfy the original equation.
x2 – 6x + 13 = 0
(3 – 2i )2 – 6(3 – 2i ) + 13 ≟ 0
9 – 12i + 4i 2 – 18 + 12i + 13 ≟ 0
4 + 4(–1) ≟ 0
0 = 0
Substitute 3 – 2i for x.
Simplify.
The left side equals the right side, so 3 – 2i checks.
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Example 3 – Solve by Completing the Square
Solve x2 = 2x + 6 by completing the square.
Solution:
x2 = 2x + 6
x2 – 2x = 6
x2 – 2x + 1 = 6 + 1
(x – 1)2 = 7
Isolate the constant term.
Complete the square.
Factor and simplify.
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Example 3 – Solution
x – 1 =
The exact solutions of x2 = 2x + 6 are and .
A calculator can be used to show that and
The decimals –1.646 and 3.646 are approximate solutions of x2 = 2x + 6.
cont’d
Apply the square root procedure.
Solve for x.
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Solving Quadratic Equations by Completing the Square
Completing the square by adding the square of half the coefficient of the x term requires that the coefficient of the x2 term be 1.
If the coefficient of the x2 term is not 1, then first multiply each term on each side of the equation by the reciprocal of the coefficient of x2 to produce a coefficient of 1 for the x2 term.
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Solving Quadratic Equations by Using the Quadratic Formula
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Solving Quadratic Equations by Using the Quadratic Formula
Completing the square for ax2 + bx + c = 0 (a 0) produces a formula for x in terms of the coefficients a, b, and c.
The formula is known as the quadratic formula, and it can be used to solve any quadratic equation.
The Quadratic Formula
If ax2 + bx + c = 0, a 0, then
As a general rule, you should first try to solve quadratic equations by factoring. If the factoring process proves difficult, then solve by using the quadratic formula.
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Example 5 – Solve by Using the Quadratic Formula
Use the quadratic formula to solve each of the following.
a. x2 = 3x + 5 b. 4x2 – 4x + 3 = 0
Solution:
a. x2 = 3x + 5
x2 – 3x – 5 = 0
Write the equation in standard form.
Use the quadratic formula.
a = 1, b = –3, c = – 5.
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Example 5 – Solution
The solutions are and .
b. 4x2 – 4x + 3 = 0
cont’d
The equation is in standard
form.
Use the quadratic formula.
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Example 5 – Solution
cont’d
a = 4, b = –4, c = 3.
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Example 5 – Solution
The solutions are and .
cont’d
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The Discriminant of a Quadratic Equation
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The Discriminant of a Quadratic Equation
The solutions of ax2 + bx + c = 0, (a 0), are given by
The expression under the radical, b2 – 4ac, is called the discriminant of the equation ax2 + bx + c = 0.
If b2 – 4ac 0, and then is a real number. If b2 – 4ac 0, then is not a real number.
Thus the sign of the discriminant can be used to determine whether the solutions of a quadratic equation are real numbers.
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The Discriminant of a Quadratic Equation
The Discriminant and the Solutions of a Quadratic Equation
The equation ax2 + bx + c = 0, with real coefficients and a 0, has as its discriminant b2 – 4ac.
If b2 – 4ac 0, then ax2 + bx + c = 0 has two distinct real solutions.
If b2 – 4ac = 0, then ax2 + bx + c = 0 has one real solution. The solution is a double solution.
If b2 – 4ac 0, then ax2 + bx + c = 0 has two distinct nonreal complex solutions. The solutions are conjugates of each other.
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Example 6 – Use the Discriminant to Determine the Number of Real Solutions
For each equation, determine the discriminant and state the number of real solutions.
a. 2x2 – 5x + 1 = 0
b. 3x2 + 6x + 7 = 0
c. x2 + 6x + 9 = 0
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Example 6 – Solution
a. The discriminant of 2x2 – 5x + 1 = 0 is
b2 – 4ac = (–5)2 – 4(2)(1)
= 17.
Because the discriminant is positive, 2x2 – 5x + 1 = 0 has two distinct real solutions.
b. The discriminant of 3x2 + 6x + 7 = 0 is
b2 – 4ac = 62 – 4(3)(7)
= –48.
Because the discriminant is negative, 3x2 + 6x + 7 = 0
has no real solutions.
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Example 6 – Solution
c. The discriminant of x2 + 6x + 9 = 0 is
b2 – 4ac = 62 – 4(1)(9)
= 0.
Because the discriminant is 0, x2 + 6x + 9 = 0 has one real solution.
cont’d
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Applications of Quadratic Equations
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Applications of Quadratic Equations
A right triangle contains one 90 angle. The side opposite the 90 angle is called the hypotenuse. The other two sides are called legs.
The lengths of the sides of a right triangle are related by a theorem known as the Pythagorean Theorem.
The Pythagorean Theorem states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the legs.
This theorem is often used to solve applications that involve right triangles.
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Applications of Quadratic Equations
The Pythagorean Theorem
If a and b denote the lengths of the legs of a right triangle and c the length of the hypotenuse, then c2 = a2 + b2.
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Example 7 – Determine the Dimensions of a Television Screen
A television screen measures 60 inches diagonally, and its aspect ratio is 16 to 9. This means that the ratio of the width of the screen to the height of the screen is 16 to 9. Find the width and height of the screen.
A 60-inch television screen with a 16:9 aspect ratio.
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Example 7 – Solution
Let 16x represent the width of the screen and let 9x represent the height of the screen.
Applying the Pythagorean Theorem gives us
(16x)2 + (9x)2 = 602
256x2 + 81x2 = 3600
337x2 = 3600
Solve for x.
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Example 7 – Solution
3.268 inches
The height of the screen is about 9(3.268) 29.4 inches, and the width of the screen is about 16(3.268) 52.3 inches.
cont’d
Apply the square root procedure. The plus-or-minus sign is not used in this
application because we know x is positive.
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Applications of Quadratic Equations
Quadratic equations are often used to determine the height (position) of an object that has been dropped or projected.
For instance, the position equation s = –16t 2 + v0t + s0 can be used to estimate the height of a projected object near the surface of Earth at a given time t in seconds.
In this equation, v0 is the initial velocity of the object in feet per second and s0 is the initial height of the object in feet.
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Other Types of Equations
SECTION 1.4
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Polynomial Equations
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Polynomial Equations
Some polynomial equations that are neither linear nor
quadratic can be solved by the various techniques
presented in this section.
For instance, the third-degree equation, or cubic
equation, in Example 1 can be solved by factoring the
polynomial and using the zero product principle.
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Example 1 – Solve a Polynomial Equation
Solve: x3 + 3x2 – 4x – 12 = 0
Solution:
x3 + 3x2 – 4x – 12 = 0
(x3 + 3x2) – (4x + 12) = 0
x2(x + 3) – 4(x + 3) = 0
(x + 3)(x2 – 4) = 0
(x + 3)(x + 2)(x – 2) = 0
x + 3 = 0 or x + 2 = 0 or x – 2 = 0
x = –3 x = –2 x = 2
The solutions are –3, –2, and 2.
Factor by grouping.
Use the zero product principle.
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96
Rational Equations
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97
Rational Equations
A rational equation is one that involves rational
expressions. The following two equations are rational
equations.
When solving a rational equation, be aware of the domain
of the equation, which is the intersection of the domains of
rational expressions.
For the first equation above, –3 and 1 are excluded as
possible values of x and are not in the domain.
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98
Rational Equations
For the second equation, –1, 1, and are excluded as possible values of x and are not in the domain.
The domain is important, as shown by trying to solve
.
We begin by noting that 3 is not in the domain of the rational expressions and then multiplying each side of the equation by x – 3.
3 is not in the domain.
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99
Rational Equations
9 + 2(x – 3) = 3x
9 + 2x – 6 = 3x
3 = x
However, the proposed solution, 3, is not in the domain, and replacing x with 3 in the original equation would require division by 0, which is not defined. Therefore, the equation has no solution.
Solve for x.
Multiply each side by x – 3, the
LCD of the denominators.
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100
Example 2 – Solve a Rational Equation
Solve.
a.
b.
c.
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101
Example 2(a) – Solution
(2x + 1) + 3(x + 4) = –2
5x + 13 = –2
5x = –15
x = –3
–3 checks as a solution.
The solution is –3.
Solve for x.
Multiply each side by x + 4, the LCD of the denominators.
–4 is not in the domain.
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102
Example 2(b) – Solution
3x(x – 2) + 4 = –4x + 12
3x2 – 6x + 4 = –4x + 12
3x2 – 2x – 8 = 0
Solve for x.
Multiply each side by x – 2.
2 is not in the domain.
cont’d
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103
Example 2(b) – Solution
(3x + 4)(x – 2) = 0
3x + 4 = 0 or x – 2 = 0
x = x = 2
checks as a solution; 2 is not in the domain and does
not check as a solution.
The solution is .
cont’d
Factor and use the zero product principle.
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104
Example 2(c) – Solution
2x(x + 4) + (x + 1)(x – 3) = (x + 4)(x – 1)
2x2 + 8x + x2 – 2x – 3 = x2 + 3x – 4
2x2 + 3x + 1 = 0
Simplify.
Multiply each side by the LCD of the denominators.
3 and –4 are not
in the domain.
cont’d
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105
Example 2(c) – Solution
(2x + 1)(x + 1) = 0
2x + 1 = 0 or x + 1 = 0
x = x = –1
and –1 check as solutions.
The solutions are –1 and .
Factor and use the zero product principle.
cont’d
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106
Radical Equations
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107
Radical Equations
Some equations that involve radical expressions can be
solved by using the following principle.
The Power Principle
If P and Q are algebraic expressions and n is a positive
integer, then every solution of P = Q is a solution of
Pn = Qn.
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108
Example 3 – Solve a Radical Equation
Solve:
Solution:
Square each side of the equation.
Simplify.
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109
Example 3 – Solution
Check:
Write the quadratic equation in standard form.
cont’d
Factor and use the zero product principle.
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110
Example 3 – Solution
The solutions check. The solution are 1 and 9.
cont’d
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111
Radical Equations
Some care must be taken when using the power principle
because the equation Pn = Qn may have more solutions
than the original equation P = Q.
As an example, consider x = 3. The only solution is the real
number 3. Square each side of the equation to produce
x2 = 9, and you get both 3 and –3 as solutions.
The –3 is called an extraneous solution because it is not a
solution of the original equation x = 3.
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112
Radical Equations
Definition of an Extraneous Solution
Any solution of Pn = Qn that is not a solution of P = Q is
called an extraneous solution. Extraneous solutions may
be introduced whenever each side of an equation is raised
to an even power.
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113
Rational Exponent Equations
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114
Rational Exponent Equations
Recall that when n is a positive even integer and
(the absolute value sign is not necessary) when n
is a positive odd integer. These results can be restated
using rational exponents.
(bn)1/n = | b |, n is a positive even integer
(bn)1/n = b, n is a positive odd integer
For instance, (x2)1/2 = | x | (n is an even integer) and
(x3)1/3 = x (n is an odd integer).
It is important to remember this when solving equations
that involve a variable with a rational exponent.
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115
Rational Exponent Equations
Here is an example that shows the details.
x2/3 = 16
(x2)1/3 = 16
[(x2)1/3]3 = 163
x2 = 163
(x2)1/2 = (163)1/2
| x | = (163)1/2
| x | = 40961/2
x = 64
Rewrite x2/3 as (x2)1/3.
Cube each side of the equation.
To take the square root, raise each side of the equation to the 1/2 power.
(x2)1/2 = | x |
Use the fact that if | x | = a(a > 0), then x = a.
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116
Rational Exponent Equations
Here is a check.
x2/3 = 16
(–64)2/3 ≟ 16
[(–64)1/3]2 ≟ 16
[–4]2 ≟ 16
16 = 16
Replace x with –64.
The solution checks.
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117
Rational Exponent Equations
x2/3 = 16
(64)2/3 ≟ 16
[(64)1/3]2 ≟ 16
[4]2 ≟ 16
16 = 16
The solutions are –64 and 64.
Replace x with 64.
The solution checks.
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118
Rational Exponent Equations
Although we could use this procedure every time we solve an equation containing a variable with a rational exponent, we will rely on a shortcut that recognizes the need for the absolute value symbol when the numerator of the rational exponent is an even integer.
Here is the solution of x2/3 = 16, using this shortcut.
x2/3 = 16
(x2/3)3/2 = 163/2
Raise each side of the equation to the 3/2 (the reciprocal of 2/3) power.
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119
Rational Exponent Equations
| x | = 64
x = 64
The solutions are –64 and 64.
Now consider x3/4 = 8. We solve this equation as
x3/4 = 8
(x3/4)4/3 = 84/3
x = 16
The solution is 16.
Because the numerator in the exponent of x2/3 is an even number, the absolute value sign is necessary.
Raise each side of the equation to the 4/3 (the reciprocal of 3/4) power.
Because the numerator in the exponent of x3/4 is an odd number, the absolute value sign is not necessary.
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120
Example 5 – Solve an Equation That Involves a Variable with a Rational Exponent
Solve.
a. 2x4/5 – 47 = 115
b. 5x3/4 + 4 = 44
Solution:
a. 2x4/5 – 47 = 115
2x4/5 = 162
x4/5 = 81
Add 47 to each side.
Divide each side by 2.
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121
Example 5 – Solution
(x4/5)5/4 = 815/4
| x | = 243
x = 243
The solutions are –243 and 243.
b. 5x3/4 + 4 = 44
5x3/4 = 40
Raise each side of the equation to the 5/4 (the reciprocal of 4/5) power.
Because the numerator in the exponent of x4/5 is an even number, use absolute value.
Subtract 4 from each side.
cont’d
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122
Example 5 – Solution
x3/4 = 8
(x3/4)4/3 = 84/3
x = 16
Substituting 16 into 5x3/4 + 4 = 44, we can verify that the solution is 16.
Raise each side of the equation to the 4/3 (the reciprocal of 3/4) power.
Because the numerator in the exponent of x3/4 is an odd number, do not use absolute value.
Divide each side by 5.
cont’d
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