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

Answer 1

Sophie Adams

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!

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Answer 2

Sophie Adams

To solve the inequality (x^3 - 11x^2 - 8x + 88 \geq 0), you need to find the intervals where the expression is greater than or equal to zero. This can be done by factoring, finding critical points, and testing intervals.

Factor the polynomial if possible.
Find the critical points by setting the expression equal to zero and solving for (x).
Test intervals determined by the critical points to determine where the expression is greater than or equal to zero.

Since factoring may not always be straightforward, you might need to use numerical methods or graphing to find approximate solutions.

Without further context on the polynomial, such as restrictions on (x) or exact values of (x) for which the inequality holds, this is the general approach.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #x^3-11x^2-8x+88&gt;=0#?

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