Roots of a cubic polynomial

[naaa00]

Homework Statement

Hello there!

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

x^3 - 10x + 18 = 0

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?

 

[kushan]

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

 

[naaa00]

Ok.

Any suggestion?

 

[Deveno]

Ok.

Any suggestion?

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.

 

[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?

...

 

[Deveno]

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?

...

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?

 

[kushan]

The function x^3 - 10x + 18 has 3 dangerous roots , which will require hours of calculation , and won't be given as homework , check if you copied it correctly

 

[vela]

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

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

=> (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?

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.

 

[kushan]

yea i kind of meant that :tongue2:

 

[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

 

[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.

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?

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

 

[kushan]

is the power 3/90 ?

 

[naaa00]

no:

(x^3)/90

Same with the others...

 

[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:

 

[naaa00]

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

 

[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?

 

[Ray Vickson]

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:

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

 

[SammyS]

Homework Statement

Hello there!

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

x^3 - 10x + 18 = 0

The Attempt at a Solution

I did the following: I rewrite 18 as

...

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^2 + 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?

OK !

Which are you trying to solve?

x³ - 10x + 18 = 0

or

x³ - 10x² + 18 = 0

 ?

 

[naaa00]

hello! Oh, my bad:

This one:

x^3 - 10x + 18 = 0

 

[kushan]

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

yea all of them are real :D

 

[kushan]

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

 

[Deveno]

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?

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.

 

[Ray Vickson]

yea all of them are real :D

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

 

[kushan]

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

 

[Ray Vickson]

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

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

 

[DuncanM]

hello! Oh, my bad:

This one:

x^3 - 10x + 18 = 0

For general reference on math topics, two of my first visits are Wikipedia and MathWorld:

http://en.wikipedia.org/wiki/Cubic_function

http://mathworld.wolfram.com/CubicFormula.html

They both provide an excellent introduction to a topic such as this one, and they also present an analytical solution to this equation. I recommend reading through these pages.

If you'd like numerical solutions, perhaps simply to confirm your own results, there are many free options (in addition to the commercial software packages available).

For example, Wolfram Alpha is well-known:

http://www.wolframalpha.com/

Less well-known are Maple's Oracles:

http://maplesoft.com/studentcenter/oracles/

(You don't have to purchase Maple, but Maple powers these online applets, which offer tools for some of the most common student tasks.)

Or there is the following, another free applet:

Cubic Equation Solver

According to this program, the roots of the equation (x^3 - 10x + 18 = 0) follow:

1.9167345798813225 + 1.0107493997715207 i
1.9167345798813225 - 1.0107493997715207 i
-3.833469159762645