## Help with equation solving and polynomial inequality

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
lia
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Joined: Sun Feb 01, 2009 3:38 am
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### Help with equation solving and polynomial inequality

Not sure how to solve this for x is the element of the real number system:
x^3 = x^2 + 4x

Not sure how to solve this inequality

x^3 = 33 < 0

stapel_eliz
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Not sure how to solve this for x is the element of the real number system:
x^3 = x^2 + 4x
Follow the usual procedure: Get everything over onto one side of the "equals" sign, factor, and then try to solve the factors.

. . . . .x3 - x2 - 4x = 0

. . . . .x(x2 - x - 4) = 0

The first factor, the x, "solves" simply enough: x = 0.

For the other factor, the x2 - 1x - 4, set this equal to zero and apply the Quadratic Formula. Since the instructions specify that they're only wanting "element of the real number system" for solutions, you would have to ignore any solutions involving a negative inside the square root, but that won't be an issue for this quadratic.

Not sure how to solve this inequality

x^3 = 33 < 0

Since thirty-three isn't less than zero, I have no idea what they want, either!

Eliz.

lia
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Joined: Sun Feb 01, 2009 3:38 am
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### Re: Help with equation solving and polynomial inequality

Thanks for the help and for the inequality I meant to write x^3 + 3x < 0 Sorry.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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for the inequality I meant to write x^3 + 3x < 0
Ah; that makes more sense!

You've got a cubic, which I'm sure you're quite used to graphing. If you think about the graph of y = x3 + 3x = x(x2 + 3), you'll recognize that there is one "real" (graphable) solution, namely at x = 0. The quadratic factor is always positive (in fact, always 3 or greater).

So you've got a cubic that crosses the x-axis exactly once. This is a positive cubic (the leading coefficient is an "understood" +1), so the graph comes up from the bottom on the left, crosses the x-axis at x = 0, and then heads up to the right. There may be a "squiggle" in the middle somewhere.

With this zero in mind, what are the two intervals for this inequality? (That is, what are the two parts of the number line where the cubic isn't zero?)

With this graph in mind, on which of those those intervals is the cubic negative? (That is, on which interval is the graph below the x-axis?)

That interval is your solution. Remember to include the point x = 0 in your solution interval, since this is an "or equal to" inequality.

Eliz.

lia
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Joined: Sun Feb 01, 2009 3:38 am
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Thank you.