How do you solve #((x+3)(x-4))/(x+5)>=0#?

1 Answer
Oct 14, 2016

Answer:

#x in (-5, -3] uu [4, oo)#

Explanation:

#f(x) = ((x+3)(x-4))/(x+5)#

Note that #f(x)# is the product of #3# terms:

#(x+3)#, #(x-4)# and #1/(x+5)#

These terms change sign when #x=-5#, #x=-3# and #x=4#. Hence #f(x)# changes sign at each of these values of #x#.

When #x=-5# the denominator is zero, so #f(x)# is undefined. So #x=-5# is not part of the solution space.

When #x=-3# or #x=4# the numerator is #0# and the inequality is satisfied. So the solution space includes #{ -3, 4 }#

The three values of #x# cut the Real line into #4# intervals in which the sign of #f(x)# does not change, namely:

#(-oo, -5)#, #(-5, -3)#, #(-3, 4)#, #(4, oo)#

For large positive values of #x# we can see that #f(x) > 0#. So the signs of the left hand side in each of these intervals is:

#(-oo, -5)": " f(x) < 0#

#(-5, -3)": " f(x) > 0#

#(-3, 4)": " f(x) < 0#

#(4, oo)": " f(x) > 0#

Remembering that #{ -3, 4 }# is part of the solution space, the complete solution space is:

#(-5, -3] uu [4, oo)#

graph{((x+3)(x-4))/(x+5) [-11.83, 8.17, -30.8, 29.2]}