Question

Solve the inequality algebraically:
x^3 - 1 (is greater than or equal to) 0

Answer 1

Set your original equation from:
\[x^{3} -1 \ge 0\]
to \[x ^{3} - 1 = 0\]
then solve for x

then plug back in \["\ge"\] of where your \["="\] should be. That will give the value of "x"

Answer 2

How would you solve an equation like x^3 + 3x^2 > 10x then?
I'm at x^3 + 3x^2 - 10x = 0, but I don't know where to go from there

Answer 3

You are going in the right direction. The trick to this equation is that you're dealing with a third degree polynomial, and lucky for you, you have "x" at every coefficient.

Now divide out a "x" from the left side of your last equation that you have derived.

Then the question is when the "x" falls true for the set right side which is " 0 ".

You will end up with one variable "x" and a quadratic.
Set each of them to equal "0" then solve.

after you get your values of "x" (there should be three values).
set them on a number line. You then pick numbers in between the value points of the "x"s that you have gotten earlier to test into the original equation. This will give you the ranges of "x" that will prove true for the equation. Ask me if you need more help.