Homework Help: Roots of a cubic polynomial

1. Feb 21, 2012

naaa00

1. The problem statement, all variables and given/known data
Hello there!

I'm trying to find the roots of the following cubic polynomial

x^3 - 10x + 18 = 0

3. The attempt at a solution
I did the following: I rewrite 18 as

18 = - (x^3 - 10x)

then I did back substitution and factored out

x^3 - 10x - x^3 + 10x = 0 or x(x^2-10) - x(x^2-10) = 0

=> (x - x)(x^2-10) = 0

=> x = sqrt(10)

I'm not sure if what I did is legit and correct. I think is wrong.

Is it valid?

Last edited: Feb 21, 2012
2. Feb 22, 2012

kushan

No it is not at all legit .
You are equating zero to zero , fundamentally wrong :\

3. Feb 22, 2012

naaa00

Ok.

Any suggestion?

4. Feb 22, 2012

Deveno

we have x3 - 10x = -18

let x = w + 10/(3w)

after doing the arithmetic, you should get a quadratic in w3, which you can then solve.

5. Feb 22, 2012

naaa00

Hello there!

To be honest, I don't understand from where does the "x = w + 10/(3w)" comes from... Could you be a little bit more explicit please?

...

Last edited: Feb 22, 2012
6. Feb 22, 2012

Deveno

it's magic!

(really, it's just a substitution)

here. i'll get you started:

(w + 10/(3w))3 - 10(w + 10/(3w) = -18, so

w3 + 3w2(10/(3w)) + 3w(100/(9w2) + 1000/(27w3) - 10w - 100/(3w) = -18

continue?

7. Feb 22, 2012

kushan

The function x^3 - 10x + 18 has 3 dangerous roots , which will require hours of calculation , and wont be given as homework , check if you copied it correctly

8. Feb 22, 2012

vela

Staff Emeritus
There's nothing wrong with setting 0 equal to 0. It's true, after all.

The problem here is that the first equation is of the form $0 \times a = 0$. Obviously, this statement is true for any a. In your case, this means x2-10 can be equal to anything, so you can't assume it's equal to 0 and solve for x.

9. Feb 22, 2012

kushan

yea i kind of meant that :tongue2:

10. Feb 22, 2012

genericusrnme

There's a cubic formula, you could try memorising that :3

And as kushan said, the roots of this are not nice at all, are you sure you copied it out correctly?
Homework problems generally have nice, friendly solutions

11. Feb 22, 2012

naaa00

Hello there!

The original problem statement is:

(x^3)/90 - (x^2)/9 + 1/5

I put the whole thing equal to zero ( for the roots), and then multiply by 90

x^3 - 10x + 18 = 0

This question was on a test. The problem also asks for local extrema, but to find them is simple.

I mean the "x = w + 10/(3w)" from where does that comes from?

Last edited: Feb 22, 2012
12. Feb 22, 2012

kushan

is the power 3/90 ?

13. Feb 22, 2012

naaa00

no:

(x^3)/90

Same with the others...

14. Feb 22, 2012

kushan

I am afraid there is no other way you need to find the roots by the cubic equation
and for your information the cube roots are
And yes they are all real
beautiful i would say :yuck:

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15. Feb 22, 2012

naaa00

My worst fear is being realized... I'm still skeptical, though...

16. Feb 22, 2012

naaa00

hello there again!

I tried the following but, again, it's a litlle bit fishy...

I started from:

x^3 - 10x = -18

=> sqrt(x^3 - 10x) = sqrt(-18)

=> sqrt[x^2 (x - 10/x)] + sqrt(-18)

=> x * sqrt{(x^2 - 10)/x)} + sqrt(-18)

I divide by sqrt(-18)

=> x * sqrt{(x^2 - 10)/-18x} + 1

Then I factor 1/x

=> 1/x * {sqrt[(x^2 - 10)/-18x] + x}

From here, I say that P(x) = 0, when x = (x^2 - 10)/-18x

Is this ok?

No, this is absurd. I don't know: I refuse to use the cubic equation...

All this is plain wrong... I guess "x" is not allowed to be as a "b" or "d" on a expression like (a + b)(c + d) = 0

Right?

Last edited: Feb 22, 2012
17. Feb 22, 2012

Ray Vickson

I don't know whether one of your roots simplifies to a real (it should), but Maple gets different results from yours:

root1 = -A/3 - 10/A and
root2 = A/6 + 5/A + I*(sqrt(3)/2)*[10/A - A/3],
where A = [243 + 3*sqrt(3561)]^(1/3), and I = sqrt(-1)

root3 = conjugate of root 2.

RGV

18. Feb 22, 2012

SammyS

Staff Emeritus
OK !

Which are you trying to solve?

x3 - 10x + 18 = 0

or

x3 - 10x2 + 18 = 0

?

19. Feb 22, 2012

naaa00

This one:

x^3 - 10x + 18 = 0

20. Feb 23, 2012

kushan

yea all of them are real :D

21. Feb 23, 2012

kushan

hey those roots which i posted earlier were for x^3 - 10x^2 + 18

22. Feb 23, 2012

Deveno

it's called "vieta's substitution" and was a break-through in the solving of the general cubic. although it bears Francois Viete's name, it was probably due to Scipione dal Ferro, and later published by Cardano.

it turns out that the roots to this cubic are "rather nasty", and a numerical method like Newton's method, would probably be faster, if an exact answer isn't needed. i would be tempted to start with an initial guess of -4.

23. Feb 23, 2012

Ray Vickson

They are not all real. Root2 has a nonzero imaginary part, as does root3. In fact, roots 2 and 3 are 1.916734580 +- 1.010749400 *I .

RGV

24. Feb 23, 2012

kushan

Ray the roots which I wrote in that post were for x^3-10x^2+18 , as OP wrote different equation

25. Feb 23, 2012

Ray Vickson

Maybe, but your remark appeared in a reply to my posting, right after the formulas I gave for the roots. Therefore, I assumed you were saying the roots I gave are all real.

RGV