Mathematics
Solving Radical Inequalities
Solving radical inequalities involves finding the values of a variable that satisfy an inequality containing radicals. To solve these inequalities, one typically isolates the radical expression, squares both sides of the inequality, and then solves for the variable. It's important to check the solutions obtained to ensure they are valid for the original inequality.
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10 Key excerpts on "Solving Radical Inequalities"
eBook - PDF- Cynthia Y. Young(Author)
- 2021(Publication Date)
- Wiley(Publisher)
• Recognize the u-substitution required to transform the equation into a simpler quadratic equation. • Recognize when a polynomial equation or an equation with rational exponents can be factored either by grouping or by first factoring out a greatest common factor. 1.4.1 Radical Equations 1.4.1 Skill Solve radical equations. 1.4.1 Conceptual Check for extraneous solutions. Radical equations are equations in which the variable is inside a radical (that is, under a square root, cube root, or higher root). Examples of radical equations follow. √ _____ x − 3 = 2 √ ______ 2x + 3 = x √ _____ x + 2 + √ ______ 7x + 2 = 6 Until now your experience has been with linear and quadratic equations. Often you can transform a radical equation into a simple linear or quadratic equation. Sometimes the transformation process yields extraneous solutions, or apparent solutions that may solve the transformed problem but are not solutions of the original radical equation. Therefore, it is very important to check your answers. 128 CHAPTER 1 Equations and Inequalities When both sides of an equation are squared, extraneous solutions can arise. For example, take the equation x = 2 If we square both sides of this equation, then the resulting equation, x 2 = 4, has two solutions: x = −2 and x = 2. Notice that the value x = −2 is not in the solution set of the original equation x = 2. Therefore, we say that x = −2 is an extraneous solution. In solving a radical equation, we square both sides of the equation and then solve the resulting equation. The solutions to the resulting equation can sometimes be extraneous in that they do not satisfy the original radical equation. STUDY TIP Extraneous solutions are common when we deal with radical equations, so remember to check your answers. EXAMPLE 1 Solving an Equation Involving a Radical Solve the equation √ _____ x − 3 = 2. Solution Square both sides of the equation. (√ _____ x − 3 ) 2 = 2 2 Simplify.
eBook - PDF- Alan Tussy, R. Gustafson(Authors)
- 2012(Publication Date)
- Cengage Learning EMEA(Publisher)
1. Isolate a radical term on one side of the equation. 2. Square both sides of the equation. 3. Solve the resulting equation. 4. Check the possible solutions in the original equation. This step is required. Strategy for Solving Radical Equations Containing Square Roots Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 8.5 Solving Radical Equations 649 EXAMPLE 2 Solve: Strategy Since the radical is already isolated on one side, we can move to Step 2 of the strategy for solving radical equations and square both sides of the equation. Why Squaring both sides will produce, on the left side, the expression that simplifies to . This step clears the equation of the radical. Solution This is the equation to solve. Use the squaring property of equality and square both sides. Simplify: and . The radical has been removed. The result is a linear equation in one variable. To solve the resulting equation, add 1 to both sides. To isolate , divide both sides by 2. Check the possible solution, 41, in the original equation. This is the original equation. Substitute 41 for . Do the multiplication within the radical. Do the subtraction within the radical. True The solution of is 41. The solution set is written . {41} 2 2 x 1 9 9 9 2 81 9 2 82 1 9 x 2 2( 41 ) 1 9 2 2 x 1 9 x x 41 2 x 82 (9) 2 81 1 2 2 x 1 2 2 2 x 1 2 x 1 81 1 2 2 x 1 2 2 ( 9 ) 2 2 2 x 1 9 2 x 1 1 2 2 x 1 2 2 2 2 x 1 9 Evaluate the left side. Do not square both sides when checking! The Language of Algebra Possible solutions also are called potential or proposed solutions.
eBook - PDFIntroductory Algebra
Concepts and Graphs 2E
- Charles P. McKeague(Author)
- 2020(Publication Date)
- XYZ Textbooks(Publisher)
8.6 Equations Involving Radicals 501 Our solution checks. It is also possible to raise both sides of an equation to powers greater than 2. We only need to check for extraneous solutions when we raise both sides of an equa- tion to an even power. Raising both sides of an equation to an odd power will not produce extraneous solutions. Solve 3 √ — 4x + 5 = 3. SOLUTION Cubing both sides, we have ( 3 √ — 4x + 5) 3 = 3 3 4x + 5 = 27 4x = 22 x = 22 __ 4 x = 11 __ 2 We do not need to check x = 11 __ 2 because we raised both sides to an odd power. B Graphing Radical Equations We end this section by looking at graphs of some equations that contain radicals. Graph y = √ — x and y = 3 √ — x . SOLUTION The graphs are shown in Figures 1 and 2. Notice that the graph of y = √ — x appears in the first quadrant only, because in the equation y = √ — x , x and y cannot be negative. The graph of y = 3 √ — x appears in Quadrants 1 and 3 because the cube root of a positive number is also a positive number, and the cube root of a negative number is a negative number. That is, when x is positive, y will be positive, and when x is negative, y will be negative. The graphs of both equations will contain the origin, because y = 0 when x = 0 in both equations. EXAMPLE 8 EXAMPLE 9 FIGURE 1 y 5 √ –5 –4 –3 –2 –1 1 2 3 4 5 –5 –4 –3 –2 –1 1 2 3 4 5 (1, 1) (4, 2) y = √ -5-4-3-2-1 1 2 3 4 5 -5 -4 -3 -2 -1 1 2 3 4 5 x y (1, 1) (4, 2) x x x y −4 Undefined −1 Undefined 0 0 1 1 4 2 9 3 16 4 502 CHAPTER 8 Roots and Radicals After reading through the preceding section, respond in your own words and in complete sentence. A. What is the squaring property of equality? B. Under what conditions do we obtain extraneous solutions to equations that contain radical expressions? C. If we have raised both sides of an equation to a power, when is it not neces- sary to check for extraneous solutions? D.
eBook - PDF- Cynthia Y. Young(Author)
- 2017(Publication Date)
- Wiley(Publisher)
ANSWER (B) the set of students who are enrolled in either a math class or a history class. (A) is the smaller set because it is all of the students who are enrolled in BOTH math and history. ▼ 1.5.2 SKILL Solve linear inequalities in one variable. 1.5.2 CONCEPTUAL Understand that a linear inequal- ity in one variable has an interval solution. 132 CHAPTER 1 Equations and Inequalities The most common mistake that occurs when solving an inequality is forgetting to change the direction of, or reverse, the inequality symbol when the inequality is multiplied or divided by a negative number. EXAMPLE 3 Solving a Linear Inequality Solve and graph the inequality 5 2 3x , 23. Solution: Write the original inequality. 5 2 3x , 23 Subtract 5 from both sides. 23x , 18 Divide both sides by 23 and reverse the inequality sign. Simplify. x . 26 Solution set: 5x | x . 266 Interval notation: 126, q2 Graph: YOUR TURN Solve the inequality 5 # 3 2 2x. Express the solution in set and interval notation, and graph. 23x 23 . 18 23 ▼ A N S W E R Solution set: 5x | x # 216 Interval notation: 12q, 214 ▼ A common mistake is using cross multiplication to solve inequalities. Cross multiplication should not be used because the expression by which you are multiplying might be negative for some values of x, and that would require the direction of the inequality sign to be reversed. EXAMPLE 4 Solving Linear Inequalities with Fractions Solve the inequality 5x 3 # 4 1 3x 2 . common mistake STUDY TIP If you multiply or divide an inequality by a negative number, remember to change the direction of the inequality sign. INEQUALITY PROPERTIES Procedures That Do Not Change the Inequality Sign 1. Simplifying by eliminating parentheses 31x 2 62 , 6x 2 x and collecting like terms. 3x 2 18 , 5x 2. Adding or subtracting the same 7x 1 8 $ 29 quantity on both sides. 7x $ 21 3. Multiplying or dividing by the 5x # 15 same positive real number. x # 3 Procedures That Change (Reverse) the Inequality Sign 1.
eBook - PDFPractical Algebra
A Self-Teaching Guide
- Bobson Wong, Larisa Bukalov, Steve Slavin(Authors)
- 2022(Publication Date)
- Jossey-Bass(Publisher)
5 LINEAR INEQUALITIES So far, we’ve worked extensively with equations. However, we often deal with situa-tions where we find values of the variable in which one expression is greater than or less than another. Fortunately, despite some important differences, the methods we use here are similar to the methods we used to solve equations. 5.1 Basic Principles of Solving Inequalities First, let’s start with a basic definition: an inequality is a statement that says that one expression is greater than or less than another expression. Like equations, inequalities must be either true or false—they can’t be both at the same time. As with equations, we often want to find the values of the variables that make the inequality true. We call these the solutions to the inequality . In Chapter 1, we used number lines to represent positive and negative numbers. Recall that positive numbers appear to the right of 0, and negative numbers appear to the left of 0. Also, any number that appears to the right of another is the greater number. We use the following inequality symbols: • > (pronounced “is greater than”) • ≥ (pronounced “is greater than or equal to,” combines the > and = symbols) • < (pronounced “is less than”) • ≤ (pronounced “is less than or equal to,” combines the < and = symbols) Here are some important points about the direction of inequalities: • The inequality symbols in statements like 6 > 2 and 5 > 3 have the same direction. • The inequality symbols in statements like 6 > 2 and 4 < 5 have the opposite direction. • If we reverse the order of the numbers, then we reverse the direction of the inequality symbol, so 3 > 1 is equivalent to 1 < 3. 111 112 PRACTICAL ALGEBRA Reading and Writing Tip The > and < symbols are easily confused. Figure 5.1 can help you remember that the inequal-ity symbols “open” in the direction of the larger number: 5 3 Figure 5.1 5 is greater than 3. Example 5.1 Determine if the inequality + 1 > − 2 is true.
eBook - ePubGRE All the Quant
Effective Strategies & Practice from 99th Percentile Instructors
- (Author)
- 2023(Publication Date)
- Manhattan Prep(Publisher)
true statement.However, while equations on the GRE tend to have only one or two values as solutions, inequalities give a whole range of values as solutions—way too many to list individually.Here’s a comparison to illustrate:Equation Inequality x + 3 = 8−3 −3x = 5 x + 3 < 8−3 −3x < 5 In the equation, there’s just one solution: x must be 5. If you tried to substitute any number other than 5 for x into the equation you would get a nonsensical answer. For example, 2 + 3 does not equal 8.However, there are many different numbers that would make the inequality true—as long as those values are less than 5. Here are just a few possible values of x for the inequality:
Because inequalities tend to have many possible solutions, this topic is commonly tested in the Select All That Apply format.Is this true? Check the math: x + 3 < 8 ? Outcome x = 2 True, so 2 is a valid solution. x = −7 True, so −7 is a valid solution. x = 6 False, so 6 is not a valid solution. Check Your Skills
- Which of the following values of x are solutions to the inequality x < 10 ?
Indicate all such values.−32.59.999
Answers can be found on page 189.Simplifying Inequalities
Many of the rules you learned with respect to equations will still apply to inequalities; you’ll learn about the few exceptions in this chapter. The rules of PEMDAS apply, and whatever you do to one side of the inequality you must still do to the other side. The goal of isolating a variable remains the same:The inequalities 2x + 6 < 12 and x < 3 provide the same information, but the second inequality is much easier to understand.Simplifying via Addition and Subtraction
With equations, if you add the same number to both sides, the equation is still true. The same principle holds true for inequalities. If you add or subtract the same number from both sides, that inequality remains true:
eBook - PDFBeginning Algebra
Connecting Concepts through Applications
- Mark Clark, Cynthia Anfinson(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
127. Simplify: 1 2 3 1 n 2 2 2 . 128. Simplify: 1 2 3 n 2 2 2 . 129. Solve the inequality: 2 5 t 1 6 2 # 8 . 130. Solve the inequality: 3 t 2 2 3 $ 2 6 . Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. S E C T I O N 8 . 4 S o l v i n g R a d i c a l E q u a t i o n s 719 Solving Radical Equations LEARNING OBJECTIVES Check solutions to radical equations. Solve radical equations. Solve applications problems involving radicals. 8.4 Checking Solutions to Radical Equations In Section 8.1, radical equations were defined as equations that have a variable under the radical. An example of a radical equation is ! x 1 8 5 16 . A solution to a radical equation is a value for the variable that, when substituted into the equation, makes the equation true 1 both sides equal 2 . Example 1 Is the value a solution? Check the given value of each variable to see whether it is a solution of the radical equation. a. ! x 1 1 5 5 for x 5 16 b. ! x 2 2 9 5 4 for x 5 2 5 c. ! x 2 3 1 7 5 6 for x 5 4 SOLUTION a. Substituting x 5 16 into ! x 1 1 5 5 yields ! x 1 1 5 5 Substitute x 5 16 into the original equation. ! 16 1 1 0 5 Simplify. 4 1 1 5 5 This is a true statement, so x 5 16 is a solution. b. Substituting x 5 2 5 into ! x 2 2 9 5 4 yields ! x 2 2 9 5 4 Substitute x 5 2 5 into the original equation. !1 2 5 2 2 2 9 0 4 Simplify. ! 25 2 9 0 4 ! 16 5 4 This is a true statement, so x 5 2 5 is a solution. c. Substituting x 5 4 into ! x 2 3 1 7 5 6 yields ! x 2 3 1 7 5 6 Substitute x 5 4 into the original equation. ! 4 2 3 1 7 0 6 Simplify. ! 1 1 7 0 6 1 1 7 ? 6 x 5 4 is not a solution.
eBook - PDFCollege Algebra
Building Skills and Modeling Situations
- Charles P. McKeague, Katherine Yoshiwara, Denny Burzynski(Authors)
- 2013(Publication Date)
- XYZ Textbooks(Publisher)
The technique we will use to solve inequalities of this type involves graphing. Suppose, for example, we want to find the solution set for the inequality x 2 − x − 6 > 0. We begin by factoring the left side to obtain (x − 3)(x + 2) > 0 We have two real numbers x − 3 and x + 2 whose product (x − 3)(x + 2) is greater than zero. That is, their product is positive. The only way the product can be positive is either if both factors, (x − 3) and (x + 2), are positive or if they are both negative. To help visualize where x − 3 is positive and where it is negative, we draw a real number line and label it accordingly: Here is a similar diagram showing where the factor x + 2 is positive and where it is negative: Drawing the two number lines together and eliminating the unnecessary num- bers, we have We can see from the preceding diagram that the graph of the solution to x 2 − x − 6 > 0 is x < −2 or x > 3 3 - - - - - - - - - - - - + + + + + Sign of x - 3 x - 3 is negative when x < 3 x - 3 is positive when x > 3 + + + + + + + + + + + + - - - - - Sign of x + 2 x + 2 is negative when x < -2 x + 2 is positive when x > -2 -2 - - - - - - - - - - + + + + + - - - - - + + + + + + + + + + -2 3 Both factors negative, their product is positive Both factors positive, their product is positive Sign of x - 3 Sign of x + 2 3 -2 144 Chapter 2 Solving Equations and Inequalities Solve for x: x 2 − 2x − 8 ≤ 0. SOLUTION We begin by factoring: x 2 − 2x − 8 ≤ 0 (x − 4)(x + 2) ≤ 0 The product (x − 4)(x + 2) is negative or zero. The factors must have opposite signs. We draw a diagram showing where each factor is positive and where each factor is negative: From the diagram, we have the graph of the solution set: −2 ≤ x ≤ 4 Solve x 2 − 6x + 9 ≥ 0. SOLUTION x 2 − 6x + 9 ≥ 0 (x − 3) 2 ≥ 0 This is a special case in which both factors are the same. Because (x − 3) 2 is always positive or zero, the solution set is all real numbers.
eBook - PDFAlgebra Teacher's Activities Kit
150 Activities that Support Algebra in the Common Core Math Standards, Grades 6-12
- Judith A. Muschla, Gary Robert Muschla, Erin Muschla-Berry(Authors)
- 2015(Publication Date)
- Jossey-Bass(Publisher)
When your students find the solution of the systems of inequalities, the solution will match one of the tans or a combination of tans. To complete this activity successfully, your students should be able to graph the solution of a linear inequality and the solution of systems of equations. They will need rulers and graph paper. Introduce the activity by providing the graph of the solution of a linear inequality, such as x + y > 5. First graph the line y = − x + 5. Then graph y > − x + 5 by making the line dashed or broken. Choose any point not on the line and shade that portion of the coordinate plane that makes the inequality true. Extend this concept to graphing the solution of a system of inequalities 1 2 0 A L G E B R A T E A C H E R ’ S A C T I V I T I E S K I T by graphing the inequalities on the same axes. The overlap of the regions is the solution to the system. Review the directions on the worksheet with your students. Note that they must draw the graphs and then match their solutions with a tan or tans. ANSWERS (1) VI (2) IV (3) VI, VII (4) I (5) II (6) VII (7) II, III Reproducibles for Section 4 follow. P O L Y N O M I A L , R A T I O N A L , E X P O N E N T I A L , A N D R A D I C A L E X P R E S S I O N S 1 2 1 Copyright © 2016 by Judith A. Muschla, Gary Robert Muschla, and Erin Muschla-Berry. Name Date Period 4–1: INTERPRETING EXPRESSIONS ------------------------------------------------------------------------------------------------------------------------------------------Understanding the following is necessary for interpreting expressions. • An expression is a variable or a combination of numbers, symbols, and/or vari-ables. Examples include 4 x , x , a + b , ab , 4(2 a + b ), and x 4 . • A term is an expression using numbers or variables, or both numbers and vari-ables, to indicate a product or quotient. Examples include 4 x , x , ab , and x 4 . • A factor is two or more numbers to be multiplied.
eBook - PDF- Jerome Kaufmann, Karen Schwitters, , , Jerome Kaufmann, Karen Schwitters(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Unless otherwise noted, all content on this page is © Cengage Learning. Chapter 2 • Equations, Inequalities, and Problem Solving 108 2 7 3 , 3 x 3 , 9 3 Divide each part by 3 2 7 3 , x , 3 The solution set is a 2 7 3 , 3 b , and its graph is shown in Figure 2.20. -2 -4 0 2 4 -7 3 Figure2.20 The distance interpretation also clarifies a property that pertains to greater than situations involving absolute value. Consider the following examples. Solve 0 x 0 . 1 and graph the solutions. Solution The number, x , must be more than one unit away from zero. Thus 0 x 0 . 1 is equivalent to x , 2 1 or x . 1 The solution set is ( 2` , 2 1) ´ (1, ` ) , and its graph is shown in Figure 2.21. -2 -4 0 2 4 Figure2.21 Solve 0 x 2 1 0 . 3 and graph the solutions. Solution The number, x 2 1 , must be more than three units away from zero. Thus 0 x 2 1 0 . 3 is equivalent to x 2 1 , 2 3 or x 2 1 . 3 Solving this disjunction yields x 2 1 , 2 3 or x 2 1 . 3 x , 2 2 or x . 4 The solution set is ( 2` , 2 2) ´ (4, ` ) , and its graph is shown in Figure 2.22. -2 -4 0 2 4 Figure2.22 Examples 8 and 9 illustrate the following general property. EXAMPLE 8 ClassroomExample Solve 0 x 0 $ 3 2 and graph the solutions. EXAMPLE 9 ClassroomExample Solve 0 x 1 2 0 $ 4 and graph the solutions. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience.
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