How do you solve $x^3-11x^2-8x+88>=0$ ?

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Answer 1 · 2016-09-04T15:21:24

The solution is $-sqrt(8) ≤ x ≤ sqrt(8)$ and $11 ≤ x$

Factor by grouping.

$=>x^2(x - 11) - 8(x - 11) ≥ 0$

$=>(x^2 - 8)(x - 11)≥ 0$

$=>x ≥ +-sqrt8, 11$

Solve by selecting test points. You will find that the interval of solution is $-sqrt(8) ≤ x ≤ sqrt(8)$ and $11 ≤ x$ .

Hopefully this helps!

How do you solve x^3-11x^2-8x+88>=0x^3-11x^2-8x+88>=0?