solve the following inequality:
(x^2(5+x)(x-5))/((x+2)(x-2)) or equal to 0

Question

solve the following inequality:
(x^2(5+x)(x-5))/((x+2)(x-2))  >or equal to 0

anonymous · Answer

@Zale101

zale101 · Answer

I'm sorry, i probably can't do it. 
@freethinker

anonymous · Answer

$$\left( \frac{ x ^{2}(5+x)(x-5) }{ (x+2)(x-2) }\ge0 \right)$$

freethinker · Answer

can you simplify?

anonymous · Answer

Honestly, I wouldnt know where to start to simplify this

freethinker · Answer

maybe because it cannot be simplified? lol

anonymous · Answer

Good answer lol

freethinker · Answer

where are the math gods when you need them?

freethinker · Answer

or goddess *

anonymous · Answer

=/ yeah

anonymous · Answer

i think it can be done u just need to test a lot of intervals

freethinker · Answer

how about you just multiply both sides by x^2-4 to get rid of the fractions?

anonymous · Answer

intervals are determined by the zeros of the numerator and the zeros of the denominator

freethinker · Answer

good thinking julian

anonymous · Answer

x = 2, x = -2 for the denominator?

anonymous · Answer

Ok so wat you need to is.. divide and multiply by (x+2) and (x-2) .. so as to make the denominator (x+2)^2 (x-2)^2 .. that is +ve always.. so you can equate it with 0.. and the numerator will be x^2(5+x)*(x-5)(x+2)(x-2).. also take x^2 to the other side.. and you'll be left with..
(5+x)(x-5)(x+2)(x-2)... get it till now?

anonymous · Answer

(5+x)(x-5)(x+2)(x-2) >=0 *

anonymous · Answer

you cant multiply with x^2-4 because you dont know if its =ve or -ve.. for x>2.. it will be -ve and the inequality will be reversed.

anonymous · Answer

+ve*

anonymous · Answer

Rnr is doing well i think that freethinker r losing some intervals

anonymous · Answer

@musicalrose get what i did there?

anonymous · Answer

yes just not sure what to do with it, to me it looks like it cancels out to being no solution

anonymous · Answer

yup. ^

anonymous · Answer

from here (5+x)(x-5)(x+2)(x-2) >=0
we just need to test the intervals
(-INF, -5)(-5,-2)(-2,2)(2,5)(5, INF)

anonymous · Answer

@freethinker its ok :) this problem just sucks

anonymous · Answer

so, if it is no solution how do I say that? my problem only asks "what is the solution" with an answer box, so I worry if there wasn't one, I would have a multiple choice.

anonymous · Answer

@julian25 i think it should be (-inf,-5][-2,2][5,inf)...
for [-5,-2] and [2,5].. the expression would be negative..

raden · Answer

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