You can make take the quadratic and make it equal to zero in order to find its roots. This will help you determine the intervals for which it will be greater than zero. Show So, for
use the quadratic formula to find
For a general form quadratic equation
you can rewrite it using its roots by using the formula
In your case, you have
Now you need to find the values of #x# that will make this greater than zero
In order for the left-hand side of this inequality to be positive, you need both terms, #(x-2)# and #(x-1)# to either be both positive or both negative. For any value of #x>2# you will get
For any value of #x<1# you will get
This means that your solution set will be #x in (-oo, 1) uu (2, +oo)#. Remember that #x=1# and #x=2# are not valid solutions for this inequality because they will make the left-hand side product equal to zero. Linear InequalitiesA linear inequalityLinear expressions related with the symbols ≤, <, ≥, and >. is a mathematical statement that relates a linear expression as either less than or greater than another. The following are some examples of linear inequalities, all of which are solved in this section:
A solution to a linear inequalityA real number that produces a true statement when its value is substituted for the variable. is a real number that will produce a true statement when substituted for the variable. Linear inequalities have either infinitely many solutions or no solution. If there are infinitely many solutions, graph the solution set on a number line and/or express the solution using interval notation. Example 1Are x=−4 and x=6 solutions to 5x+7< 22? Solution: Substitute the values in for x, simplify, and check to see if we obtain a true statement.
Answer: x=−4 is a solution and x=6 is not. All but one of the techniques learned for solving linear equations apply to solving linear inequalities. You may add or subtract any real number to both sides of an inequality, and you may multiply or divide both sides by any positive real number to create equivalent inequalities. For example: 10>−510 −7>−5−7Subtrac t7onbothsides.3>−12 ✓True 10>−5105>−55 Dividebothsidesby5.2>−1 ✓True Subtracting 7 from each side and dividing each side by positive 5 results in an inequality that is true. Example 2Solve and graph the solution set: 5x+7<22. Solution: 5x+ 7<225x+7−7<22−75x<15 5x5<155x<3 It is helpful to take a minute and choose a few values in and out of the solution set, substitute them into the original inequality, and then verify the results. As indicated, you should expect x=0 to solve the original inequality and that x=5 should not.
Checking in this manner gives us a good indication that we have solved the inequality correctly. We can express this solution in two ways: using set notation and interval notation. {x| x<3}Setnotation(−∞,3)Intervalnotation In this text we will choose to present answers using interval notation. Answer: (−∞ ,3) When working with linear inequalities, a different rule applies when multiplying or dividing by a negative number. To illustrate the problem, consider the true statement 10>−5 and divide both sides by −5. 10>−510−5>−5−5Dividebothsidesb y−5.−2>1✗False Dividing by −5 results in a false statement. To retain a true statement, the inequality must be reversed. 10> −510−5<−5−5 Reversetheinequality.−2<1✓True The same problem occurs when multiplying by a negative number. This leads to the following new rule: when multiplying or dividing by a negative number, reverse the inequality. It is easy to forget to do this so take special care to watch for negative coefficients. In general, given algebraic expressions A and B, where c is a positive nonzero real number, we have the following properties of inequalitiesProperties used to obtain equivalent inequalities and used as a means to solve them.:
We use these properties to obtain an equivalent inequalityInequalities that share the same solution set., one with the same solution set, where the variable is isolated. The process is similar to solving linear equations. Example 3Solve and graph the solution set: −2(x+8)+6≥20. Solution: −2(x+8)+6≥ 20Distribute.− 2x−16+6≥20Combine liketerms.−2x−10≥20 Solveforx.−2x≥30 Dividebothsidesby−2.−2x −2≤30−2Reversetheinequality.x≤−15 Answer: Interval notation (−∞,−15] Example 4Solve and graph the solution set: −2(4x−5)<9−2(x−2). Solution: −2(4x−5)<9−2(x−2)−8x+10<9 −2x+4−8x+10<13−2x−6x<3−6x− 6>3−6 Reversetheinequality.x>−12 Answer: Interval notation (−12,∞) Example 5Solve and graph the solution set: 12x−2≥12(74x−9)+1. Solution: 12x−2≥12(74x−9)+112x−2≥78x−92+112x −78x≥−72+2−38x≥−32 (−83)(−38x)≤ (−83)(−32)R eversetheinequality. x≤4 Answer: Interval notation: ( −∞,4] Try this! Solve and graph the solution set: 10−5(2x+3)≤25. Answer: [−3,∞); Compound InequalitiesFollowing are some examples of compound linear inequalities:
These compound inequalitiesTwo or more inequalities in one statement joined by the word “and” or by the word “or.” are actually two inequalities in one statement joined by the word and or by the word or. For example, −13<3x−7<17 is a compound inequality because it can be decomposed as follows: −13<3x−7and3x−7<17 We can solve each inequality individually; the intersection of the two solution sets solves the original compound inequality. While this method works, there is another method that usually requires fewer steps. Apply the properties of this section to all three parts of the compound inequality with the goal of isolating the variable in the middle of the statement to determine the bounds of the solution set. Example 6Solve and graph the solution set: −13<3x−7<17. Solution: −13<3x−7<17−13 +7<3x−7+7<17+7−6<3x<24−63< 3x3<243−2<x<8 Answer: Interval notation: (−2,8) Example 7Solve and graph the solution set: 56≤13(1 2x+4)<2. Solution: 56≤13(12x+4)<2 56≤16x+43<26⋅(56)≤6⋅(16x+43 )<6⋅(2)5≤x+8<125−8≤x+8−8<12−8 −3≤x<4 Answer: Interval notation [−3,4) It is important to note that when multiplying or dividing all three parts of a compound inequality by a negative number, you must reverse all of the inequalities in the statement. For example: −10<−2x<20−10−2>−2x−2>20−25>x> −10 The answer above can be written in an equivalent form, where smaller numbers lie to the left and the larger numbers lie to the right, as they appear on a number line. −10<x<5 Using interval notation, write: ( −10,5). Try this! Solve and graph the solution set: −3≤−3(2x−3)<15. Answer: (−1,2]; For compound inequalities with the word “or” you work both inequalities separately and then consider the union of the solution sets. Values in this union solve either inequality. Example 8Solve and graph the solution set: 4x+5≤−15or 6x−11>7. Solution: Solve each inequality and form the union by combining the solution sets.
Answer: Interval notation (−∞,−5]∪(3,∞) Try this! Solve and graph the solution set: 5(x−3)<−20or2(5−3x )<1. Answer: (−∞,−1)∪(32,∞); Applications of Linear InequalitiesSome of the key words and phrases that indicate inequalities are summarized below:
As with all applications, carefully read the problem several times and look for key words and phrases. Identify the unknowns and assign variables. Next, translate the wording into a mathematical inequality. Finally, use the properties you have learned to solve the inequality and express the solution graphically or in interval notation. Example 9Seven less than 3 times the sum of a number and 5 is at most 11. Find all numbers that satisfy this condition. Solution: First, choose a variable for the unknown number and identify the key words and phrases. Let n represent the unknown indicated by “a number.” Solve for n. 3(n+5)−7≤11 3n+15−7≤113n+8≤113n≤3n ≤1 Answer: Any number less than or equal to 1 will satisfy the statement. Example 10To earn a B in a mathematics course the test average must be at least 80% and less than 90%. If a student earned 92%, 96%, 79%, and 83% on the first four tests, what must she score on the fifth test to earn a B? Solution: Set up a compound inequality where the test average is between 80% and 90%. In this case, include the lower bound, 80. Let x represent the score on the fifth test. 80≤ testaverage<9080≤92+96+79+83 +x5<905⋅80≤5⋅350+x5<5⋅90400≤350+x<45050≤x<100 Answer: She must earn a score of at least 50% and less than 100%. In the previous example, the upper bound 100% was not part of the solution set. What would happen if she did earn a 100% on the fifth test? average=92+96+79+83+1005=4505 =90 As we can see, her average would be 90%, which would earn her an A. Key Takeaways
Topic Exercises
Part A: Linear InequalitiesDetermine whether or not the given value is a solution. Graph all solutions on a number line and provide the corresponding interval notation.
Part B: Compound InequalitiesGraph all solutions on a number line and provide the corresponding interval notation.
Part C: ApplicationsFind all numbers that satisfy the given condition. Set up an algebraic inequality and then solve.
Part D: Discussion BoardAnswers
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