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How do you solve #(x+8)^2(x+5)(x+7)^2>=0#?

1 Answer

Jul 2, 2016

#x in { -8, -7 } uu [-5, oo)#

Explanation:

Both #(x+8)^2# and #(x+7)^2# are squares, so always non-negative for Real values of #x#.

As a result, the left hand side is negative when #x+5 < 0# unless #x+7 = 0# or #x+8=0#, when it is zero.

Hence the inequality is satisfied when: #x in { -8, -7 } uu [-5, oo)#

graph{(x+8)^2(x+5)(x+7)^2 [-13.96, 6.04, -7.68, 2.32]}

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