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## Homework Statement

Find the zeros of the cubic equation:

[tex]y = x^3 -9x^2 + 15x + 30[/tex]

How do we find the zeros of this? In this case, subbing in x-values that will make it equal 0 does not work.

- Thread starter aeromat
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Find the zeros of the cubic equation:

[tex]y = x^3 -9x^2 + 15x + 30[/tex]

How do we find the zeros of this? In this case, subbing in x-values that will make it equal 0 does not work.

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Char. Limit

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vela

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Ruffini's method:

http://en.wikipedia.org/wiki/Ruffini's_rule

http://en.wikipedia.org/wiki/Ruffini's_rule

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gb7nash

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Char. Limit

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If he's restricting himself to real numbers, there ARE no other roots. Just the one.I would probably suggest using Newton's method (or bisection method if you can find a positive function value and a negative function value) Once you've found the first root, take f(x)/(x-r), where r is the root. This will yield a quadratic and you can use the quadratic formula to find the remaining two roots.

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gb7nash

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I'm missing something here. How do we know there's only one real root?

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Char. Limit

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Oh, sorry. I graphed it first, to get an impression of where the roots were.

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gb7nash

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Cheater. :tongue:

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Char. Limit

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Real mathematicians use pictures!

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Dick

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You can also conclude there is only one root by looking at the derivative and finding the extreme values plus knowing the behavior as x->+/-infinity. Which is basically 'graphing it' without a calculator. Hence, not cheating.I'm missing something here. How do we know there's only one real root?

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gb7nash

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Good call. This is a good method.You can also conclude there is only one root by looking at the derivative and finding the extreme values plus knowing the behavior as x->+/-infinity. Which is basically 'graphing it' without a calculator. Hence, not cheating.

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Char. Limit

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Well, Newton's method is given as such: given a guess x_{0} as to where the root of a function f(x) is, then...

[tex]x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}[/tex]

x_{1} is a better guess, and x_{2} will be a better guess than x_{1}, and so on.

NOTE: With an initial guess close to the answer, you can get a very good approximation after only two or three iterations of this. For example, in this problem, an initial guess of x_{0}=-1 will get you a very good approximation in three iterations.

[tex]x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}[/tex]

x

NOTE: With an initial guess close to the answer, you can get a very good approximation after only two or three iterations of this. For example, in this problem, an initial guess of x

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Dick

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If you don't know numerical techniques and you are good with a calculator or computer program there is nothing wrong with bisection either. If f(x)=x^3−9x^2+15x+30 then f(-2) is negative and f(0) is positive. That means there is a root in between. Check the mean at (-2+0)/2=(-1). Depending on the sign of f(-1) keep splitting the interval in half until you get whatever precision you want. You'll never get it exact, but do you need to?

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