How do you solve #(x+3)/(x-8)<0#?

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Answer 1 · 2017-07-22T11:22:48

Solution: # -3 < x <8 # , in interval notation: #(-3,8)#

# (x+3)/(x-8) < 0 ; x !=8 # critical points are #x=8 , x = -3#

Sign Change:

For # x < -3 # , sign of # (x+3)/(x-8)# is #(-)/(-) = + i.e >0#

For # -3 < x <8 # , sign of # (x+3)/(x-8)# is #(+)/(-) = - i.e <0#

For # x > 8 # , sign of # (x+3)/(x-8)# is #(+)/(+) = + i.e >0#

So, Solution: # -3 < x <8 # , in interval notation: #(-3,8)#

The graph also confirms above findings.

graph{(x+3)/(x-8) [-11.25, 11.25, -5.625, 5.625]} [Ans]