What is the formula to find the Zeros of this?

[konartist]

What is the formula to find the Zeros of this?

$$f(x)= x^3 - 3x^2 - 9x + 9$$

I just can't seem to find them.
Rational root theorem:
+-1,3,9
(nope)
won't work with synthetic substitution

Can't group it either

What da hell?

 

[Pengwuino]

I just ran it through mathematica and the roots are complex

 

[J77]

I just ran it through mathematica and the roots are complex

Not according to my quick Matlab look - they don't look easy to find tho'...

4.54461,-2.37755,0.83294[/color]

 

[HallsofIvy]

Well, the certainly can't all be complex!

A quick graph shows there are 3 real roots, one just less than -2, one just less than +1, one around 4.5.

There is a cubic formula but it is a lot more complicated than the quadratic formula. See:
http://www.math.vanderbilt.edu/~schectex/courses/cubic/

 

[Curious3141]

The cubic is fairly easy to solve (the quartic's the killer).

Don't memorise the formula, BTW. It's a lot easier to learn the principle of the thing. You can find it here : http://www.karlscalculus.org/cubic.html

 

[J77]

I wonder if the OP wrote down the correct equation. Students shouldn't have to use the cubic equation to find roots, ie., one root should be easy to factor out, leaving a quadratic to solve - none of the roots are easy to find (by inspection) in the case here.

 

[Curious3141]

I wonder if the OP wrote down the correct equation. Students shouldn't have to use the cubic equation to find roots, ie., one root should be easy to factor out, leaving a quadratic to solve - none of the roots are easy to find (by inspection) in the case here.

When I was in school, we had these chemical equilibrium questions to do. Now I never liked Chem a whole lot so I liked to fool around. Often to make these questions solvable at the kiddie level (junior high school I suppose), simple approximations such as assuming the concentration of a particular ion in abudance would not change would be made. That way the system could be reduced to solving a linear, or at the worst, a quadratic equation.

To get the teacher's goat, I purposely didn't use to make those approximations, instead choosing to solve the problem exactly. This used to give me cubics most of the time, and occasional quartics. We weren't "supposed" to know how to do these, but I did them anyway with all the complicated and tedious working (hey, I had a lot of time to kill ). Needless to say, it pissed the teacher off a good deal. Of course, they couldn't mark me wrong, since the method and the answer were, if anything, more correct than the ones they expected us to be using.

 

[konartist]

I just learned the Newton Method today, but it still isn't an exact zero. I really don't think there is anyway to solve this analyticaly, maybe just by a calculator?

EDIT: I have mathematica 5.0, how do I enter this into the program?? Is there a retard's guide to mathematica?

 

[Curious3141]

I just learned the Newton Method today, but it still isn't an exact zero. I really don't think there is anyway to solve this analyticaly, maybe just by a calculator?

We've told you there's a perfectly well known and not too difficult way to solve it analytically should you choose to, and I've provided a reference. The question is, do you want to ?

Yes, Mathematica can solve this equation exactly in radicals. I've been out of touch with it, but there's a Solve[] function built in that gives exact solutions where possible. Perhaps someone else can help you with this.

 

[konartist]

Oh, ok! I didn't see the link there.

Well, if anybody can tell me what to plug into mathematica I'd like to check my answers.

 

[Curious3141]

Oh, ok! I didn't see the link there.

Well, if anybody can tell me what to plug into mathematica I'd like to check my answers.

Umm..why ?

If you want to solve it numerically, then you don't need Mathematica. Just substitute the solutions you find back into the equation with a calculator and verify the RHS comes very close to zero (within the limit of round-off error).

If you want to solve it analytically by hand, then you still don't need Mathematica. Just work it out in terms of radicals or trig ratios, then punch in the solution into your scientific calc to get a decimal number. Then put that back into the equation and again verify that it's a valid solution to the equation within round-off error.

The only thing Mathematica can help you with is to solve the equation analytically in radicals. To verify that, you can use a simple calc.

If you have 3 real roots, there's no problem at all verifying the roots. If you a single real root and a pair of complex conjugate roots, you can use a late model sci. calc. with complex number algebraic functions to do the trick.

 

[Curious3141]

Anyway, if you're interested, I've solved the equation exactly by hand. Took 10 minutes, I recommend you try it.

The solution set is given by :

$$x = 1 + 4\cos(\frac{1}{3}\arctan\sqrt{63} + \frac{2}{3}k\pi)$$

where k takes the values 0, 1, 2 in succession to give 3 real roots. Don't forget to work in radians.