Intermediate Algebra. 7.6 Quadratic Inequalities. Name. Problem Set 7.6 Solutions to Every Odd-Numbered Problem. Date
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1 7.6 Quadratic Inequalities 1. Factoring the inequality: x 2 + x! 6 > 0 ( x + 3) ( x! 2) > 0 The solution set is x <!3 or x > 2. Graphing the solution set: 3. Factoring the inequality: x 2! x! 12 " 0 ( x + 3) ( x! 4) " 0 The solution set is!3 " x " 4. Graphing the solution set: 5. Factoring the inequality: x 2 + 5x! "6 x 2 + 5x + 6! 0 ( x + 2) ( x + 3)! 0 The solution set is x! "3 or x # "2. Graphing the solution set:
2 7. Factoring the inequality: 6x 2 < 5x! 1 6x 2! 5x + 1 < 0 ( 3x! 1) ( 2x! 1) < 0 The solution set is 1 3 < x < 1. Graphing the solution set: 2 9. Factoring the inequality: x 2! 9 < 0 ( x + 3) ( x! 3) < 0 The solution set is!3 < x < 3. Graphing the solution set: 11. Factoring the inequality: 4x 2! 9 " 0 ( 2x + 3) ( 2x! 3) " 0 The solution set is x! " 3 2 or x # 3. Graphing the solution set: 2
3 13. Factoring the inequality: 2x 2! x! 3 < 0 ( 2x! 3) ( x + 1) < 0 The solution set is!1 < x < 3. Graphing the solution set: Factoring the inequality: x 2! 4x + 4 " 0 ( x! 2) 2 " 0 Since this inequality is always true, the solution set is all real numbers. Graphing the solution set: 17. Factoring the inequality: x 2! 10x + 25 < 0 ( x! 5) 2 < 0 Since this inequality is never true, there is no solution. 19. The solution set is 2 < x < 3 or x > 4. Graphing the solution set:
4 21. The solution set is x! "3 or " 2! x! "1. Graphing the solution set: 23. The solution set is!4 < x " 1. Graphing the solution set: 25. Combining the fractions and simplifying: 3x x + 6! 8 x + 6 < 0 3x! 8 x + 6 < 0 The solution set is!6 < x < 8. Graphing the solution set: 3
5 27. Combining the fractions and simplifying: 4 x! > x! 6 > 0 x! 6 x! 2 x! 6 > 0 The solution set is x < 2 or x > 6. Graphing the solution set: 29. The solution set is x <!3 or 2 < x < 4. Graphing the solution set: 31. Combining the fractions and simplifying: 2 x! 4! 1 x! 3 > 0 2x! 6! x + 4 ( x! 4) x! 3 ( ) > 0 x! 2 ( ) > 0 ( x! 4) x! 3 The solution set is 2 < x < 3 or x > 4. Graphing the solution set:
6 33. Combining the fractions and simplifying: x + 7 2x x 2! 36 " 0 x x + 6 x + 6 ( ) + 6 ( )( x! 6) " 0 ( x + 7) ( x! 6) + 12 " 0 2( x + 6) ( x! 6) x 2 + x! x + 6 ( )( x! 6) " 0 x 2 + x! 30 2 x + 6 ( )( x! 6) " 0 ( x + 6) ( x! 5) 2( x + 6) ( x! 6) " 0 x! 5 2( x! 6) " 0 The solution set is 5! x < 6. Graphing the solution set: 35. a. The solution set is!2 < x < a. The solution set is!2 < x < 5. b. The solution set is x <!2 or x > 2. b. The solution set is x <!2 or x > 5. c. The solution set is x =!2,2. c. The solution set is x =!2, a. The solution set is x <!1 or 1 < x < 3. b. The solution set is!1 < x < 1 or x > 3. c. The solution set is x =!1,1, 3.
7 41. Let w represent the width and 2w + 3 represent the length. Using the area formula: w 2w + 3 ( )! 44 2w 2 + 3w! 44 2w 2 + 3w " 44! 0 ( 2w + 11) ( w " 4)! 0 The width is at least 4 inches. 43. Solving the inequality: 1300 p! 100 p 2 " 4000!100 p p! 4000 " 0 p 2! 13p + 40 # 0 ( p! 8) ( p! 5) # 0 Charge at least $5 but no more than $8 per radio. 45. Let x represent the number of $10 increases in dues. Then the revenue is given by: y = 10000! 200x ( )( x) =!2000x x ( ) =!2000 x 2! 40x =!2000 x! 20 ( ) 2 + 1,800,000 The dues should be increased by $200, so the dues should be $300 to result in a maximum income of $1,800,000.
8 47. Let x represent the number of $2 increases in price. Then the income is given by: y = 40! 2x ( )( x) =!4x x ( ) =!4 x 2! 10x + 25 ( ) =!4 x! 5 The owner should make five $2 increases in price, resulting in a price of $ Using a calculator: 50,000 32,000 = Using a calculator: 1 " # $ ! 2 % & ' ( Solving the equation: 54. Solving the equation: 3t! 1 = 2 x + 3 = x! 3 ( 3t! 1) 2 = ( 2) ( 2 x + 3) 2 = ( x! 3) 2 3t! 1 = 4 x + 3 = x 2! 6x + 9 3t = 5 0 = x 2! 7x + 6 t = 5 0 = ( x! 6) ( x! 1) 3 x = 1,6 x = 1 does not check The solution is 5. The solution is Graphing the equation: ( )
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