Question

If x a set of all real numbers R the find the solution set for $(2x-3)(x+8)<0$

Answer 1

Case 1 :
When (2x-3) > 0 and (x+8) <0
i.e. x > 3/2 and x < -8
This is not possible as as a number greater than 3/2 cannot be less than -8

Case 2:
When (2x-3) < 0 and (x+8) > 0
i.e. x < 3/2 and x > -8
-8 < x < 3/2, which is possible.
Therefore Solution set = $(-8,3/2)$