INTRODUCTORY EXERCISES

(2) Solve each inequality.

 (a) (x + 3)(x - 4)  0

 (b) 9/4  x2

 (c) 3x2 - 2x - 6 < 2x2 - 6x - 1

 (d) x - 2
     ------ > 0
     2x + 5

 (e)   2x
     ------ < 0
     x2 + 5

 (a) (x+3)(x-4)  0

     either both factors are positive or both are negative

     let (x+3)  0 and (x-4)  0
         x  -3        x  4
     Therefore x  4

     now let (x+3)  0 and (x-4)  0
             x  -3        x  4
     Therefore we get that x  -3

     The solution is (-,-3)(4,)


 (b) 9/4  x2

     3/2   x

     x  3/2  or  x  -3/2


 (c) 3x2 - 2x - 6 < 2x2 - 6x - 1

     x2 + 4x - 5 < 0
     (x+5)(x-1) < 0
     
     first we try when
     x + 5 < 0       and       x - 1 > 0
     x < -5          and       x > 1    
       impossible situation.

     now try
     x + 5 > 0       and       x - 1 < 0
     x > -5          and       x < 1

     The answer is -5 < x < 1


 (d) x - 2
     ------ > 0
     2x + 5

     Assume both the numerator and denominator are positive, then
     x - 2 > 0    and    2x + 5 > 0
     x > 2        and    2x > -5
                         x > -5/2

     Now assume both are negative, we get:
     x - 2 < 0    and    2x + 5 < 0
     x < 2        and    x < -5/2

     Hence we get the solution to be (-,-5/2)  (2,)


 (e)   2x
     ------ < 0
     x2 + 5

     Since x2+5 is ALWAYS greater than zero, we just have to see where
     2x < 0
     x < 0