14. ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 2 2 3 3 ⎜ ⎜ ⎟ ⎜ ⎟ Interval −∞, − − , ∞ , + 7 + 12 < 0 x2 x ⎝ ⎠ ⎝ ⎠ 2 3 2 ⎠ 3 = + + ( 7 12 f x) x2 x Test −1 0 2 Number + 7 + 12 = x2 x + −6 7 Value of 5 8 + + f 0 ( 4)( 3) x x x Conclusion Positive Negative Positive = 0 < ⎧ 2 3 ⎫ =− = −3 4, are the zeros of x x f . The solution set is ⎨ x ⎬ or, − < < x ⎩ ⎭ 3 2 ⎛ 2 3 ⎞ interval notation, ⎜ − , ⎟ . ⎝ 3 2 ⎠ − > 7) x( x } 8 − < < −3 4 or, x 17. { The solution set is x − 7 > x2 x using 8 (−4, − 7 − 8 > 0 x2 x interval notation, −3) . 15. ( = − 7 − 8 f x) x2 x 2 < 5x + 2 x2 3 − 7 − 8 = 0 x x 2 − 5x − 3 < 0 x2 2x2 − 5x − 3 = 0 (2 + 1)( − 3) = 0 x x = − − ( 2 5x f x) x2 3 + − = ( 1)( 8) 0 x x = −1, = 8 are the zeros of . x x f 1 , = 3 are the zeros of x x f 2 { } −1 The solution set is < or > 8 or, using x x x (−∞, (8, ) −1) ∪ ∞ interval notation, . + 1) > 20 x 18. + > 20 x2 x The solution set is ⎧ x 2 − 1 < < 3 ⎫ or, using x ⎨ 2 ⎬ + − 20 > 0 x x ⎩ ⎭ interval notation, ⎛ − 1 , 3 ⎞ . = + − ( 20 f x) x2 x 2 ⎜ ⎟ 2 + − 20 = 0 x x ( = 6 − 5x − 6 f x) x2 ⎝ ⎠ 6x2 − 5x − 6 = 0 (3x + 2)(2 − 3) = 0 x 16. 6 < 6 + 5x x2 ( + 5)( − 4) = 0 x x − − 6 5x 6 < 0 x2
14. + 7 x < x2 + 7 x + 12 < 0 f ( x) = x2 + 7 x + 12 x2 + 7 x + 12 = 0 ( x + 4)( x + 3) = 0 x =− 4, x = −3 are the zeros of f . Interval ⎛ 2 ⎞ ⎜ −∞, − 3 ⎠ ⎛ 2 3 ⎞ ⎜ − , ⎟ ⎝ 3 2 ⎠ ⎛ 3 ⎞ ⎜ , ∞ ⎟ ⎝ 2 ⎠ Test Number −1 0 2 Value of f 5 −6 8 Conclusion Positive Negative Positive ⎧ 2 3 ⎫ The solution set is ⎨ x − < x < ⎬ or, ⎩ 3 2 ⎭ ⎛ 2 3 ⎞ interval notation, ⎜ − , ⎟ . The solution set is { x − 4 < x < −3 } or, using 17. x( x − 7) > 8 x2 − 7 x > 8 ⎝ 3 2 ⎠ 15. interval notation, (−4, −3) . 2 x2 < 5x + 3 x2 − 7 x − 8 > 0 f ( x) = x2 − 7 x − 8 2 x − 7 x − 8 = 0 2 x2 − 5x − 3 < 0 f ( x) = 2 x2 − 5x − 3 2x2 − 5x − 3 = 0 (2 x + 1)( x − 3) = 0 ( x + 1)( x − 8) = 0 x = −1, x = 8 are the zeros of f . x 1 , x = 3 are the zeros of f 2 The solution set is { x x < −1 or x > 8 } or, using interval notation, (−∞, −1) ∪ (8, ∞ ) . The solution set is ⎧ x − 1 < x < 3 ⎫ or, using 18. x + 1) > 20 x2 + x > 20 2 ⎨ 2 ⎬ x + x − 20 > 0 ⎩ ⎭ interval notation, ⎛ − 1 , 3 ⎞ . f ( x) = x2 + x − 20 2 ⎜ 2 ⎟ x + x − 20 = 0 16. ⎝ ⎠ 6 x2 < 6 + 5x 6 x2 − 5x − 6 < 0 f ( x) = 6 x2 − 5x − 6 6x2 − 5x − 6 = 0 (3x + 2)(2 x − 3) = 0 ( x + 5)( x − 4) = 0