How can you solve cubic equations in R with multiple variables?

[santa]

solve in R

(x^2+2)^{1/3}+(4x^2+3x-2)^{1/3}=(3x^2+x+5)^{1/3}+(2x^2+2x-5)^{1/3}

 

[EnumaElish]

Any thoughts?

 

[VietDao29]

solve in R

(x^2+2)^{1/3}+(4x^2+3x-2)^{1/3}=(3x^2+x+5)^{1/3}+(2x^2+2x-5)^{1/3}

You should notice that:
(x² + 2) + (4x² + 3x - 2) = (3x² + x + 5) + (2x² + 2x - 5) = 5x² + 3x.

So, if you let: \alpha = x ^ 2 + 2
\beta = 3x ^ 2 + x + 5
and \gamma = 5x ^ 2 + 3x

Then, your original equation will becomes:

\sqrt[3]{\alpha} + \sqrt[3]{\gamma - \alpha} = \sqrt[3]{\beta} + \sqrt[3]{\gamma - \beta}

When \gamma = 0, both sides equal 0, so this is one obvious solution. You'll get 2 (of a toal 4) solutions when setting \gamma = 0, and solve for x.

On graphing it, I can see there are actually 4 solutions.. Not sure how you can obtain another 2 solutions.

 

[sadhu]

it has got only one solution i.e 0

 

[CompuChip]

Then, your original equation will becomes:

\sqrt[3]{\alpha} + \sqrt[3]{\gamma - \alpha} = \sqrt[3]{\beta} + \sqrt[3]{\gamma - \beta}

When \gamma = 0, both sides equal 0

Are you sure that
\sqrt[3]{\alpha} + \sqrt[3]{-\alpha} = \sqrt[3]{\alpha - \alpha} = 0

 

[VietDao29]

Are you sure that
\sqrt[3]{\alpha} + \sqrt[3]{-\alpha} = \sqrt[3]{\alpha - \alpha} = 0

We are working in the reals, aren't we?

Btw, it's not like that, since: \sqrt[3]{a} + \sqrt[3]{b} \neq \sqrt[3]{a + b}.

It should've read:

\sqrt[3]{\alpha} + \sqrt[3]{-\alpha} = \sqrt[3]{\alpha} - \sqrt[3]{\alpha} = 0[/QUOTE]

:)

 

[CompuChip]

Yes, we are working in the reals, and the identity I gave obviously doesn't hold.. that's what confused me.
I still don't really get what you try to do. Setting \gamma = 0 means that
\gamma = 5 x^2 + 3 x = x(5 x + 3) = 0
so x = 0 or x = -3/5.
Taking x = 0 the formula becomes
\sqrt[3]{2} + \sqrt[3]{-2} = \sqrt[3]{5} + \sqrt[3]{-5}
which, numerically, is something like
1.89 + 1.09 i = 2.56 + 1.48 i

So it appears I didn't get what you were trying to say...

Instead, there appear to be two real (non-zero) solutions and two complex solutions (which we are not interested in). I have no idea how to find them though.

 

[VietDao29]

Taking x = 0 the formula becomes
\sqrt[3]{2} + \sqrt[3]{-2} = \sqrt[3]{5} + \sqrt[3]{-5}
which, numerically, is something like
1.89 + 1.09 i = 2.56 + 1.48 i

Well, no. We are not working in the complex. "i" belongs to the complex, not the reals. In the complex, \sqrt[3]{a} has 3 different values (for a <> 0), so actually \sqrt[3]{a} + \sqrt[3]{b} has at most 6 different values.

Whereas, in the reals, \sqrt[3]{a} only has one value, i.e, the real number whose cubed is a.

E.g, we have: \sqrt[3]{8} + \sqrt[3]{-8} = 2 + (-2) = 0 (since (-2)³ = -8)

In general, we have: \sqrt[3]{-a} = -\sqrt[3]{a}.

Proof:

Let b = \sqrt[3]{a}, b is unique, since we are in the reals.
\Rightarrow b ^ 3 = a (the definition of cube root)
\Rightarrow - b ^ 3 = -a (take the additive inverse of both sides)
\Rightarrow (- b) ^ 3 = -a
\Rightarrow - b = \sqrt[3]{-a} (again, the definition of cube root).
So we have \sqrt[3]{-a} = -\sqrt[3]{a} (Q.E.D)

And so, that means:

\sqrt[3]{2} + \sqrt[3]{-2} = \sqrt[3]{2} - \sqrt[3]{2} = 0 = \sqrt[3]{5} + \sqrt[3]{-5}

Is everything clear now? :)

 

[santa]

good work but

let $ \sqrt[3]{x^2+2}=a, \ \sqrt[3]{4x^2+3x-2}=b, \ \sqrt[3]{3x^2+x+5}=c, \ \sqrt[3]{2x^2+2x-5}=d.

these may be help

 

[VietDao29]

good work but

let $ \sqrt[3]{x^2+2}=a, \ \sqrt[3]{4x^2+3x-2}=b, \ \sqrt[3]{3x^2+x+5}=c, \ \sqrt[3]{2x^2+2x-5}=d.

these may be help

Then, how will you go from there??

 

[NateTG]

On graphing it, I can see there are actually 4 solutions.. Not sure how you can obtain another 2 solutions.

\alpha=\beta
and
\gamma=\alpha + \beta

Are obviously solutions - no idea if that gets you anything.

Edit: Looks like those are all imaginary solutions.

 

[VietDao29]

\alpha=\beta
and
\gamma=\alpha + \beta

Are obviously solutions - no idea if that gets you anything.

Edit: Looks like those are all imaginary solutions.

Well, ya, why didn't I think of it before. Damn it. >.<

\alpha = \beta will gives you 2 solutions in the complex, whereas \gamma = \alpha + \beta will give you another 2 real solutions as wishes.

If you are allowed to graph, and then solve for x, then it's done. Woohoo. But if you are not allowed to graph :(, then you must first prove that it has at most 4 solutions, then point out all the 4 solutions.

But well, I think, in this problem, you are allowed to use graph.

 

[Rainbow Child]

solve in R

(x^2+2)^{1/3}+(4x^2+3x-2)^{1/3}=(3x^2+x+5)^{1/3}+(2x^2+2x-5)^{1/3}

Let

\alpha=x^2+2,\,\beta=4x^2+3x-2\,\gamma=3x^2+x+5,\,\delta=2x^2+2x-5

with \alpha\neq 0,\,\beta\neq 0,\,\gamma \neq 0,\delta\neq 0 since the roots of these equations does not satisfy the original equation. Then

\alpha^{1/3}+\beta^{1/3}=\gamma^{1/3}+\delta^{1/3}\neq 0 \quad (1) and
\alpha+\beta=\gamma+\delta \quad (2)

Raising (1) to the 3rd power

\alpha+3\,\alpha^{2/3}\,\beta^{1/3}+3\,\alpha^{1/3}\,\beta^{2/3}+\beta=\gamma+3\,\gamma^{2/3}\,\delta^{1/3}+3\,\gamma^{1/3}\,\delta^{2/3}+\delta\Rightarrow

3\,\alpha^{2/3}\,\beta^{1/3}+3\,\alpha^{1/3}\,\beta^{2/3}=3\,\gamma^{2/3}\,\delta^{1/3}+3\,\gamma^{1/3}\,\delta^{2/3}\Rightarrow \quad \text{from (2)}

\alpha^{1/3}\,\beta^{1/3}\,(\alpha^{1/3}+\beta^{1/3})=\gamma^{1/3}\,\delta^{1/3}\,(\gamma^{1/3}+\delta^{1/3})\Rightarrow

\alpha^{1/3}\,\beta^{1/3}=\gamma^{1/3}\,\delta^{1/3}\Rightarrow \quad \text{from (1)}

\alpha\,\beta=\gamma\,\delta \quad (3)

Thus we have to solve the system of (1), (2), (3) which is rather simple. From (3) we have

\frac{\alpha}{\gamma}=\frac{\delta}{\beta}=k

which makes (2)

(k-1)\,\gamma=(k-1)\,\beta

with k\neq 1 because leads to inconsistency the original equation. Thus \gamma=\beta and from (2) \alpha=\delta, with the final answer

x=-1-2\,\sqrt{2},\, x=-1+2\,\sqrt{2}​

P.S. I hope this time the mentors would not delete this post, after the conversation we had about santa's threads.