Question

If x a set of all real numbers R the find the solution set for $(x+3)(x-5)<0$

Answer 1

Case 1 :
When (x+3) < 0 and (x-5) >0
i.e. x < -3 and x > 5
This is not possible as as a number less than -3 cannot be greater than 5

Case 2:
When (x+3) > 0 and (x-5) < 0
i.e. x > -3 and x < 5
-3 < x < 5, which is possible.
Therefore Solution set = $(-3,5)$