Section 1.6 Inequalities

Before Starting this Section, Review

  1. 1 Inequalities (Section P.1 , pages 6 and 7)

  2. 2 Intervals (Section P.1 , page 8)

  3. 3 Linear equations (Section 1.1 , page 83)

Objectives

  1. 1 Learn the vocabulary for discussing inequalities.

  2. 2 Solve and graph linear inequalities.

  3. 3 Solve and graph a compound inequality.

  4. 4 Solve polynomial and rational inequalities using test points.

The Bermuda Triangle

The Bermuda Triangle, or Devil’s Triangle, is the expanse of the Atlantic Ocean between Florida, Bermuda, and Puerto Rico covering approximately 500,000 square miles of sea. This mysterious stretch of sea has an unusually high occurrence of disappearing ships and planes.

For example, on Halloween 1991, pilot John Verdi and his copilot were flying a Grumman Cougar jet over the triangle. They radioed the nearest tower to get permission to increase their altitude. The tower agreed and watched as the jet began the ascent and then disappeared off the radar. The jet did not fly out of range of the radar, did not descend, and did not radio a mayday (distress call). It just vanished, and the plane and crew were never recovered.

One explanation for such disappearances is based on the theory that strange compass readings occur in crossing the Atlantic due to confusion caused by the three north poles: magnetic (toward which compasses point), grid (the real North Pole, at 90 degrees latitude), and true or celestial north (determined by Polaris, the North Star). To get data to test this theory, a plane set out from Miami to Bermuda, a distance of 1035 miles, with the intention of setting the plane on automatic pilot once it reached a cruising speed of 300 miles per hour.

The plane was 150 miles along its path from Miami to Bermuda when it reached the 300-mile-per-hour cruising speed and was set on automatic pilot. After what length of time would you be confident in saying that the plane had encountered trouble? The answer to this question is found in Example 2, which uses a linear inequality, the topic discussed in this section.

Inequalities

  1. 1 Learn the vocabulary for discussing inequalities.

When we replace the equal sign (=)(=) in an equation with any of the four inequality symbols <, ≤, >, or ≥, the resulting expression is an inequality. Here are some examples.

Equation Replace=with Inequality
x=53x+2=145x+7=3x+23x2=0 <≤>≥ x<53x+2≤145x+7>3x+23x2≥0

In general, an inequality is a statement that one algebraic expression is either less than or is less than or equal to another algebraic expression. The vocabulary for discussing inequalities is similar to that used for equations. The domain of a variable in an inequality is the set of all real numbers for which both sides of the inequality are defined. The real numbers that result in a true statement when those numbers are substituted for the variable in the inequality are called solutions of the inequality. Because replacing x with 1 in the inequality 3x+2≤14 results in the true statement

3(1)+2≤14,

the number 1 is a solution of the inequality 3x+2≤14. We also say that 1 satisfies the inequality 3x+2≤14. To solve an inequality means to find all solutions of the inequality — that is, its solution set. The most elementary equations, such as x=5, have only one solution. However, even the most elementary inequalities, such as x<5, have infinitely many solutions. In fact, their solution sets are intervals, and we frequently graph the solution sets for inequalities in one variable on a number line. The graph of the inequality x<5 is the interval (−∞, 5), shown in Figure 1.6.

Figure 1.6

Graph of an inequality

Inequalities are classified the same way as equations: conditional, inconsistent, or identities. A conditional inequality such as x<5 has in its domain at least one solution and at least one number that is not a solution. An inequality that no real number satisfies is called an inconsistent inequality, and an inequality that is satisfied by every real number in the domain of the variable is called an identity.

Because x2=x⋅x is the product of (1) two positive factors, (2) two negative factors, or (3) two zero factors, x2 is either a positive number or zero. That is, x2 is never negative, or is nonnegative. We call this fact the nonnegative identity.

Two inequalities that have exactly the same solution set are called equivalent inequalities. The basic method of solving inequalities is similar to the method for solving equations: We replace a given inequality with a series of equivalent inequalities until we arrive at an equivalent inequality, such as x<5, whose solution set we already know.

Caution is required in multiplying or dividing both sides of an inequality by a real number or an expression representing a real number. Notice what happens when we multiply an inequality by −1.

We know that 2<3, but how does −2=(−1)(2) compare with −3=(−1)(3)? We have −2>−3. See Figure 1.7.

Figure 1.7

Side Note

On the number line a<b corresponds to the point a being to the left of the point b.

If we multiply (or divide) both sides of the inequality 2<3 by −1, to get a correct result, we must exchange the < symbol for the > symbol. This exchange of symbols is called reversing the sense or the direction of the inequality.

The following chart describes the way multiplication and division affect inequalities.

Linear Inequalities

  1. 2 Solve and graph linear inequalities.

A linear inequality in one variable is an inequality that is equivalent to one of the forms

ax+b<0orax+b≤0,

where a and b represent real numbers and a≠0.

Inequalities such as 2x−1>0 and 2x−1≥0 are linear inequalities because they are equivalent to −2x+1<0 and −2x+1≤0, respectively.

Side Note

A linear inequality becomes a linear equation when the inequality symbol is replaced with the equal sign (=).

Example 1 Solving and Graphing Linear Inequalities

Solve each inequality and graph its solution set.

  1. 2x<−2

  2. 7x−11<2(x−3)

  3. 8−3x≥2

Solution

  1. 2x<−2

    x<−1Divide both sides by 2.

    The solution set is {x|x<−1} or, in interval notation, (−∞, −1). The graph is shown in Figure 1.8.

    Figure 1.8

  2. 7x−11<2(x−3)

    7x−11<2x−6Distributive property7x−11+11<2x−6+11Add 11 to both sides.7x<2x+5Simplify.7x−2x<2x+5−2xSubtract 2x from both sides.5x<5Simplify.5x5<55Divide both sides by 5.x<1Simplify.

    The solution set is {x|x<1} or, in interval notation, (−∞, 1). The graph of the solution set is shown in Figure 1.9.

    Figure 1.9

  3. 8−3x≥2

    8−3x−8≥2−8Subtract 8 from both sides.−3x≥−6Simplify.−3x−3≤−6−3Divide both sides by −3. (Reverse thedirection of the inequality symbol.)x≤2Simplify.

    The solution set is {x|x≤2} or, in interval notation, (−∞, 2]. The graph of the solution set is shown in Figure 1.10.

    Figure 1.10

Practice Problem 1

  1. Solve each inequality and graph the solution set.

    1. 3x+1≤10

    2. 4x+9>2(x+6)+1

    3. 7−2x≥−3

Example 2 Calculating the Results of the Bermuda Triangle Experiment

In the introduction to this section, we discussed an experiment to test the reliability of compass settings and flight by automatic pilot along one edge of the Bermuda Triangle. The plane is 150 miles along its path from Miami to Bermuda, cruising at 300 miles per hour, when it notifies the tower that it is now set on automatic pilot.

The entire trip is 1035 miles, and we want to determine how much time we should let pass before we become concerned that the plane has encountered trouble.

Solution

Let t=time elapsed since the plane went on autopilot. Then

300t=distance the plane has flown in t hours150+300t=plane’s distance from Miami after t hours.

Concern is not warranted unless the Bermuda tower does not see the plane, which means that the following is true.

Plane's distance from Miami≥Distance from Miami to Bermuda150+300t≥1035Replace the verbal description withan inequality.150+300t−150≥1035−150Subtract 150 from both sides.300t≥885Simplify.300t300≥885300Divide both sides by 300.t≥2.95Simplify.

Because 2.95 is roughly three hours, the tower will suspect trouble if the plane has not arrived in Bermuda after three hours.

Practice Problem 2

  1. How much time should pass in Example 2 if the plane was set on automatic pilot at 340 miles per hour when it was 185 miles from Miami?

If an inequality is true for all real numbers, its solution set is (−∞, ∞). If an inequality has no solution, its solution set is ∅.

Example 3 Solving an Inequality

Write the solution set of each inequality.

  1. 7(x+2)−20−4x<3(x−1)

  2. 2(x+5)+3x<5(x−1)+3

Solution

  1. 7(x+2)−20−4x<3(x−1)Original inequality7x+14−20−4x<3x−3Distributive property3x−6<3x−3Combine like terms.−6<−3Add −3x to both sides and simplify.

    The last inequality is equivalent to the original inequality and is always true. So the solution set of the original inequality is (−∞, ∞).

  2. 2(x+5)+3x<5(x−1)+3Original inequality2x+10+3x<5x−5+3Distributive property5x+10<5x−2Combine like terms.10<−2Add −5x and simplify.

    The resulting inequality is equivalent to the original inequality and is always false. So the solution set of the original inequality is ∅.

Practice Problem 3

  1. Write the solution set of each inequality.

    1. 2(4−x)+6x<4(x+1)+7

    2. 3(x−2)+5≥7(x−1)−4(x−2)

Combining Two Inequalities

  1. 3 Solve and graph a compound inequality.

Sometimes we are interested in the solution set of two or more inequalities. The combination of two or more inequalities is called a compound inequality.

Suppose E1 is an inequality with solution set (interval) I1 and E2 is another inequality with solution set I2. Then the solution set of the compound inequality “E1 and E2” is I1∩I2 and the solution set of the compound inequality “E1 or E2” is I1∪I2.

Example 4 Solving and Graphing a Compound OR Inequality

Graph and write the solution set of the compound inequality.

2x+7≤1or3x−2<4(x−1)

Solution

2x+7≤1or3x−2<4(x−1)Original inequalities2x+7≤1or3x−2<4x−4Distributive property2x≤−6or−x<−2Isolate variable term.x≤−3orx>2Solve for x;reverse secondinequality symbol.

Graph each inequality and select the union of the two intervals.

x≤−3

x>2

x≤−3orx>2

The solution set of the compound inequality is (−∞, −3]∪(2, ∞).

Practice Problem 4

  1. Write the solution set of the compound inequality.

    3x−5≥7or5−2x≥1

Example 5 Solving and Graphing a Compound AND Inequality

Solve the compound inequality and graph the solution set.

2(x−3)+5<9and3(1−x)−2≤7

Solution

2(x−3)+5<9and3(1−x)−2≤7Original inequalities2x−6+5<9and3−3x−2≤7Distributive property2x<10and−3x≤6≤6Isolate variable term.x<5andx≥−2Solve for x; reverse secondinequality symbol.

Graph each inequality and select the interval common to both inequalities.

x<5

x≥−2

x<5andx≥−2

The solution set of the compound inequality is the interval [−2, 5).

Practice Problem 5

  1. Solve and graph the compound inequality.

    2(3−x)−3<5and2(x−5)+7≤3

Sometimes we are interested in a joint inequality such as −5<2x+3≤9. This use of two inequality symbols in a single expression is shorthand for −5<2x+3 and 2x+3≤9. Fortunately, solving such inequalities requires no new principles, as we see in the next example.

Example 6 Solving and Graphing a Compound Inequality

Solve the inequality −5<2x+3≤9 and graph its solution set.

Solution

We must find all real numbers that are solutions of both inequalities

−5<2x+3and2x+3≤9.

We first solve these inequalities separately.

−5<2x+32x+3≤9Original inequalities−5−3<2x+3−32x+3−3≤9−3Subtract 3 from both sides.−8<2x2x≤6Simplify.−82<2x22x2≤62Divide both sides by 2.−4<xx≤3Simplify.

The solution of the original pair of inequalities consists of all real numbers x such that −4<x and x≤3. This solution may be written more compactly as {x|−4<x≤3}. In interval notation, we write (−4, 3]. The graph is shown in Figure 1.11.

Figure 1.11

Notice that we did the same thing to both inequalities in each step of the solution process. We can accomplish this simultaneous solution more efficiently by working on both inequalities at the same time, as follows.

−5<2x+3≤9Original inequalities−5−3<2x+3−3≤9−3Subtract 3 from both sides.−8<2x≤6Simplify.−82<2x2≤62Divide both sides by 2.−4<x≤3Simplify.

The solution set is (−4, 3], the same solution set we found previously.

Practice Problem 6

  1. Solve and graph −6≤4x−2<4.

Example 7 Finding the Interval of Values for a Linear Expression

If −2<x<5, find real numbers a and b so that a<3x−1<b.

Solution

We start with the interval for x.

−2<x<53(−2)<3x<3(5)Multiply each part by 3 to get 3x in the middle.−6<3x<15Simplify.−6−1<3x−1<15−1Subtract 1 from each part to get 3x−1 in the middle.−7<3x−1<14Simplify.

We have a=−7 and b=14.

Practice Problem 7

  1. Assuming that −3≤x≤2, find real numbers a and b so that a≤3x+5≤b.

Our next example demonstrates a practical use of the method shown in Example 7.

Example 8 Finding a Fahrenheit Temperature from a Celsius Range

The weather in London is predicted to range between 10° and 20° Celsius during the three-week period you will be working there. To decide what kind of clothes to take, you want to convert the temperature range to Fahrenheit temperatures. The formula for converting Celsius temperature C to Fahrenheit temperature F is F=95C+32. What range of Fahrenheit temperatures might you find in London during your stay there?

Solution

First, we express the Celsius temperature range as an inequality. Let C=temperature in Celsius degrees.

For the three weeks under consideration, 10≤C≤20.

We want to know the range of F=95C+32 when 10≤C≤20.

10≤C≤20(95)(10)≤95C≤(95)(20)Multiply each part by 95.18≤95C≤36Simplify.18+32≤95C+32≤36+32Add 32 to each part.50≤95C+32≤68Simplify.50≤F≤68F=95C+32

So the temperature range from 10° to 20° Celsius corresponds to a range from 50° to 68° Fahrenheit.

Practice Problem 8

  1. What range in Fahrenheit degrees corresponds to the range of 15° to 25° Celsius?

Using Test Points to Solve Inequalities

  1. 4 Solve polynomial and rational inequalities using test points.

The test-point method, also known as the sign-chart method, involves writing an inequality (by rearranging if necessary) so that the expression on the left side of the inequality symbol is in factored form and the right side is 0. In Example 9, we illustrate this method to solve a quadratic inequality.

Practice Problem 9

  1. Solve: x2+2<3x+6

A rational inequality is an inequality that includes one or more rational expressions. We use the test-point method to solve a rational inequality.

Example 10 Solving a Rational Inequality

Solve: x2+2x−15x−1≥3

  1. Step 1  x2+2x−15x−1−3≥0Rearrange so that the right side is 0.x2+2x−15x−1−3(x−1)x−1≥0Common denominator(x2+2x−15)−3(x−1)x−1≥0ac±bc=a±bcx2+2x−15−3x+3x−1≥0Distribute.x2−x−12x−1≥0Simplify.

  2. Step 2 (x+3)(x−4)x−1≥0Factor the numerator.

  3. Step 3

    x+3=0for x=−3x−4=0for x=4x−1=0for x=1The rational expression is undefined at x=1.

    The four intervals determined by the three points −3, 1, and 4 are

    (−∞, −3), (−3, 1), (1, 4), and (4, ∞).

  4. Step 4

    Test Interval Test Point Sign of (x+3)(x−4)(x−1) Result
    (−∞, −3) −4 (−)(−)(−) Negative
    (−3, 1)   0 (+)(−)(−) Positive
    (1, 4)   2 (+)(−)(+) Negative
    (4, ∞)   5 (+)(+)(+) Positive

  5. Step 5

  6. Step 6 (x+3)(x−4)(x−1)≥0 on the set [−3, 1)∪[4, ∞), with the following graph:

Practice Problem 10

  1. Solve: 2x+5x−1≤1

Warning

You cannot solve a rational inequality by multiplying both sides by the LCD, as you would with a rational equation. Remember that you reverse the sense of an inequality when multiplying by a negative expression. However, when an expression contains a variable, you do not know whether the expression is positive or negative.

Section 1.6 Exercises

Concepts and Vocabulary

In Exercises 1–4, fill in the blank with the correct inequality symbol, using the rules for producing equivalent inequalities.

  1. If x<8, then x−8_0.

  2. If x≤−3, then x+3_0.

  3. If x2<6, then x_12.

  4. If x3<−2, then x_−6.

  5. True or False. If −2x≤4, then 2x≥−4.

  6. True or False. One solution of the inequality −3x≥12 is −4.

  7. True or False. One solution of the inequality x≥7 is 7.001.

  8. True or False. Every real number is a solution of the inequality −5x≤x.

Building Skills

In Exercises 9–18, graph the solution set of each inequality and write it in interval notation.

  1. −2<x<5

  2. −5≤x≤0

  3. 0<x≤4

  4. 1≤x<7

  5. x≥−1

  6. x>2

  7. −5x≥10

  8. −2x<2

  9. 2x−3>−5

  10. 5x+1≤16

In Exercises 19–40, solve each inequality. Write the solution in interval notation and graph the solution set.

  1. x+3<6

  2. x−2<3

  3. 1−x≤4

  4. 7−x>3

  5. 2x+5<9

  6. 3x+2≥7

  7. 3−3x>15

  8. 8−4x≥12

  9. 3(x+2)<2x+5

  10. 4(x−1)≥3x−1

  11. 3(x−3)≤3−x

  12. −x−2≥x−10

  13. 6x+4>3x+10

  14. 4(x−4)>3(x−5)

  15. 8(x−1)−x≤7x−12

  16. 3(x+2)+2x≥5x+18

  17. 5(x+2)≤3(x+1)+10

  18. x−4>2(x+8)

  19. 2(x+1)+3≥2(x+2)−1

  20. 5(2−x)+4x≤12−x

  21. 2(x+1)−2≤3(2−x)+9

  22. 4(1−x)+2x>5(2−x)+4x

In Exercises 41–50, solve each rational inequality.

  1. 9x−6≥32x+9

  2. 7x−32<3x−4

  3. x−33≤2+x2

  4. 2x−34≥3−x2

  5. 3x+12<x−1+x2

  6. 2x−13≥x+14+x12

  7. x−32≥x3+1

  8. 2x+13<x−12+16

  9. 3x+13−x2≤x+22

  10. x−13+x+14≤x2+x12

In Exercises 51–58, solve each compound or inequality.

  1. 2x+5<1 or 2+x>4

  2. 3x−2>7 or 2(1−x)>1

  3. 2x−34≤2 or 4−3x2≥2

  4. 5−3x3≥16 or x−13≤1

  5. 2x+13≥x+1 or x2−1>x3

  6. x+22<x3+1 or x−13>x+15

  7. x−12>x3−1 or 2x+53≤x+16

  8. 2x+13≤x4+1 or 3−x2>x3−1

In Exercises 59–66, solve each compound and inequality.

  1. 3−2x≤7 and 2x−3≤7

  2. 6−x≤3x+10 and 7x−14≤3x+14

  3. 2(x+1)+3≥1 and 2(2−x)>−6

  4. 3(x+1)−2≥4 and 3(1−x)+13>4

  5. 2(x+1)−3>7 and 3(2x+1)+1<10

  6. 5(x+2)+7<2 and 2(5−3x)+1<17

  7. 5+3(x−1)<3+3(x+1) and 3x−7≤8

  8. 2x−3>11 and 5(2x+1)<3(4x+1)−2(x−1)

In Exercises 67–78, solve each compound inequality.

  1. 3<x+5<4

  2. 9≤x+7≤12

  3. −4≤x−2<2

  4. −3<x+5<4

  5. −9≤2x+3≤5

  6. −2≤3x+1≤7

  7. 0≤1−x3<2

  8. 0<5−x2≤3

  9. −1<2x−35≤0

  10. −4≤5x−23≤0

  11. 5x≤3x+1<4x−2

  12. 3x+2<2x+3<4x−1

In Exercises 79–84, find a and b.

  1. If −2<x<1, then a <x+7<b.

  2. If 1<x<5, then a <2x+3<b.

  3. If −1<x<1, then a <2−x<b.

  4. If 3<x<7, then a <1−3x<b.

  5. If 0<x<4, then a <5x−1<b.

  6. If −4<x<0, then a <3x+4<b.

In Exercises 85–100, use the test-point method to solve each polynomial inequality.

  1. x2+4x−12≤0

  2. x2−8x+7>0

  3. 6x2+7x−3≥0

  4. 4x2−2x−2<0

  5. (x+3)(x+1)(x−1)≥0

  6. (x+4)(x−1)(x+2)≤0

  7. x3−4x2−12x>0

  8. x3+8x2+15x>0

  9. x2+2x<−1

  10. 4x2+12x<−9

  11. x3−x2≥0

  12. x3−9x2≥0

  13. x2≥1

  14. x4≤16

  15. x3<−8

  16. x4>9

In Exercises 101–120, solve each rational inequality.

  1. x+2x−5<0

  2. x−3x+1>0

  3. x+4x<0

  4. xx−2>0

  5. x+1x+2≤3

  6. x−1x−2≥3

  7. (x−2)(x+2)x>0

  8. (x−1)(x+3)x−2<0

  9. (x−2)(x+1)(x−3)(x+5)≥0

  10. (x−1)(x−3)(x+2)(x+4)≥0

  11. x2−1x2−4≤0

  12. x2−9x2−64≤0

  13. x+43x−2≥1

  14. 2x−3x+3≤1

  15. 3≤2x+62x+1

  16. x−22x+1<−1

  17. x+2x−3≥x−1x+3

  18. x+1x−2≥xx−1

  19. x−1x+1≤x+2x−3

  20. x+3x+1≤x−1x−2

Applying the Concepts

  1. Appliance markup. The markup over the dealer’s cost on a new refrigerator ranges from 15% to 20%. If the dealer’s cost is $1750, over what range will the selling price vary?

  2. Return on investment. An investor has $5000 to invest for a period of one year. Find the range of per annum simple interest rates required to generate interest between $200 and $275 inclusive.

  3. Hybrid car trip. Sometime after passing a truck stop 300 miles from the start of her trip, Cora’s hybrid car ran out of gas. Assuming that the tank could hold 12 gallons of gasoline and the hybrid car averaged 40 miles per gallon, find the range of gasoline (in gallons) that could have been in the tank at the start of the trip.

  4. Average grade. Sean has taken three exams and earned scores of 85, 72, and 77 out of a possible 100 points. He needs an average of at least 80 to earn a B in the course. What range of scores on the fourth (and last) 100-point test will guarantee a B in the course?

  5. Butterfat content. How much cream that is 30% butterfat must be added to milk that is 3% butterfat to have 270 quarts that are at least 4.5% butterfat?

  6. Pedometer cost. A company produces a pedometer at a cost of $3 each and sells the pedometer for $5 each. If the company has to recover an initial expense of $4000 before any profit is earned, how many pedometers must be sold to earn a profit in excess of $3000?

  7. Car sales. A car dealer has three times as many SUVs and twice as many convertibles as four-door sedans. How many four-door sedans does the dealer have if she has at least 48 cars of these three types?

  8. Temperature conversion. The formula for converting Fahrenheit temperature F to Celsius temperature C is C=59(F−32). What range in Celsius degrees corresponds to a range of 68° to 86° Fahrenheit?

  9. Temperature. The number N of water mites in a water sample depends on the temperature t in degrees Fahrenheit and is given by N=132t−t2. At what temperature will the number of mites exceed 3200?

  10. Falling object. The height h of an object thrown from the top of a ski lift 1584 feet high after t seconds is h=−16t2+32t+1584. For what times is the height of the object at least 1200 feet?

  11. Blackjack. A gambler playing blackjack in a casino using four decks noticed that 20% of the cards that had been dealt were jacks, queens, kings, or aces. After x cards had been dealt, he knew that the likelihood that the next card dealt would be a jack, a queen, a king, or an ace was 64−0.2x208−x. For what values of x is this likelihood greater than 50%?

  12. Area of a triangle. The base of a triangle is 3 centimeters greater than the height. Find the possible heights h so that the area of such a triangle will be at least 5 square centimeters.

Beyond the Basics

  1. Find the numbers k for which the quadratic equation 2x2+kx+2=0 has two real solutions.

  2. Find the numbers k for which the quadratic equation 2x2+kx+2=0 has no real solutions.

  3. Find the numbers k for which the quadratic equation x2+kx+k=0 has two real solutions.

  4. Find the numbers k for which the quadratic equation x2+kx+k=0 has no real solutions.

  5. Solve: x2x+1≥14 and 6x4x−1<12

  6. Solve: 2x−1x−7>1 and x−10x−8>2

  7. Find all values of c for which it is possible to find two numbers whose product is 36 and whose sum is c.

  8. Find all values of c for which it is possible to find two numbers whose sum is 12 and the sum of whose squares is c.

  9. An import firm pays a tax of $10 on each radio it imports. In addition to this tax, a penalty tax must be paid if more than 1000 radios are imported. The penalty tax is computed by multiplying 5 cents by the number of radios imported in excess of 1000. This tax must be paid on each radio imported. If 1006 radios are imported, the penalty tax is 6⋅5=30 cents and is paid on each of the 1006 radios. If the firm wants to spend no more than a total of $640,000 on import taxes, how many radios can it import?

  10. A TV quiz program pays a contestant $100 for each correct answer for ten questions. If all ten questions are answered correctly, bonus questions are asked. The reward for every correct answer is increased by $50 for each bonus question that is correctly answered. Any incorrect answer ends the game. If two bonus questions are answered correctly, the contestant receives $200 for each of the 12 questions. If a contestant won more than $3500, how many questions must have been answered correctly?

Critical Thinking/Discussion/Writing

  1. Give an example of a quadratic inequality with each of the following solution sets.

    1. (−4, 5)

    2. [−2, 6]

    3. (−∞, ∞)

    4. ∅

    5. {3}

    6. (−∞, 2)∪(2, ∞)

  2. Give an example of an inequality with each of the following solution sets.

    1. (−2, 4]

       

    2. [3, 5)

       

    3. Is there a quadratic inequality whose solution set is (2, 5]?

Getting Ready for the Next Section

In Exercises 145–150, evaluate each expression.

  1. |−3|

  2. |3−7|

  3. |6−4|

  4. |−√2|

  5. |0|

  6. |−15.8|

In Exercises 151–154, find the distance between the given pair of points on a number line.

  1. −2 and 5

  2. −8 and −15

  3. 2.3 and 5.7

  4. −5 and 0

In Exercises 155–160, express each statement in algebraic notation using absolute values.

  1. The distance from x to −2 is 5.

  2. x is either 3 or −3.

  3. x is at most 2 units from 4.

  4. x is either to the left of −5 or to the right of 5 on a number line.

  5. x is within 3 units of 5.

  6. x is closer to 2 than to 6.

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