How do you solve #x^ { 3} + 6x ^ { 2} - 36x \leq 216#?

1 Answer
mradul tiwari
Mar 25, 2017

x #<=#6 is the solution.

Explanation:

#x^3 + 6x^2 - 36x -216<=0#
check by hit and trial, put x=6 we get L.H.S = 0, so (x-6) is a factor of given equation.
now, given can be written as:
#x^2(x-6) +12x(x-6)+36(x-6) <=0 rArr (x-6)(x^2+12x+36)<=0 rArr (x-6)((x+6)^2)<=0 rArrx<=6#
now use wavy curve method to solve the inequality. if you don't know that than google it.
graph{x<=6 [-10, 10, -5, 5]} #

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