## almost homework, a system in 3 unknowns

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alessandro95
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### almost homework, a system in 3 unknowns

I was doing a very easy homework question (that I have solved in the "normal" way) when I thought of an alternative solution but I got stuck when I found myself facing a system of 3 equations in 3 unknowns since I have no idea on how to deal with it (unless the equations are linear, but those are not).

The 3 equations are:
-2(x+y+z)=3
xy+xz+yz=3
xyz=6

for my excercise I need to show that any triple (x,y,z) solving the system contains at most one real number.
Since I've never dealt with something like this before I tried to show that there are no triples of real numbers that satisfies all the equations but I couldn't find a simple way to show it. (well, of course, I can just paste them into wolfram alpha but I was supposed to provide a pen&paper solution).

I'm pretty sure that it is easy to show that there are no 3 real numbers solving the system, but I don't know if showing that every triple solving it contains at most one real is even feasible.

Every help (including keywords to google or stuff to read) would be very appreciated!
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Dopefish
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### Re: almost homework, a system in 3 unknowns

Isn't the traditional route just taking one of your equations, solve it for one of the unknowns (e.g. x=6/(yz)), plug that into another equation which now has only 2 unknowns, solve for one of those in terms of the other, and then plug that into the final equation to solve for a value?

Then backtrack using that value to solve for the other two, which would presumably spit out some imaginary numbers, although I haven't done it.

You can swap around the labels without changing anything in any of your equations too so it seems clear that once you get one solution (x,y,z) then any permutation of that would also be a solution, although I don't know how rigorous that is.

jaap
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### Re: almost homework, a system in 3 unknowns

Dopefish wrote:Isn't the traditional route just taking one of your equations, solve it for one of the unknowns (e.g. x=6/(yz)), plug that into another equation which now has only 2 unknowns, solve for one of those in terms of the other, and then plug that into the final equation to solve for a value?

You could do that, but it gets very messy.
There is however a particular insight that makes this problem fairly straightforward. Here are some hints.
Spoiler:
Think about cubic polynomials.

Spoiler:
Cubics always have at least one real valued root.

Spoiler:
What are the coefficients of the cubic you get when you expand (x-r1)(x-r2)(x-r3) ?

alessandro95
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### Re: almost homework, a system in 3 unknowns

the original excercise was about a polynomial and his roots so I obtained those equations from a cubic polynomial, following your suggestion would bring me back where I started
The primary reason Bourbaki stopped writing books was the realization that Lang was one single person.

jaap
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### Re: almost homework, a system in 3 unknowns

Well, doing the whole messy substitution business that dopefish suggested would probably also eventually get you back to the original cubic polynomial equation. It is the simplest equation that x satisfies (as well as y or z), and as far as I can tell, using the cubic is the simplest way to get to the answers.

Schrollini
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### Re: almost homework, a system in 3 unknowns

If you can show that there is no solution of three reals that satisfy those equations (I don't immediately see this), then the rest is fairly straightforward. As jaap points out, all cubics have at least one real solution, so all you have to do is prove that you can't have two real numbers and one complex number. WLOG, let x be real and y be complex. Then from your first equation, you immediate have the complex part of z, so you know whether z will be real or complex.

But this does require you to show that at least one of the variables in complex.
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PM 2Ring
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### Re: almost homework, a system in 3 unknowns

Um, perhaps we should be a little more explicit and say that all cubics with real coefficients have at least one real solution.

alessandro95, are you familiar with Cardano's & Vieta's methods for solving cubics?
http://en.wikipedia.org/wiki/Cubic_function

I guess it's also worth mentioning that
(u + v)³ = 3uv(u + v) + (u³ + v³)
We can use that to solve a cubic of the form
x³ = ax + b
by letting x = u+v, and then solving
a = 3uv & b = u³ + v³
And we can transform any cubic to a form without an x² term by a simple substitution, as the Wikipedia link shows.

eta oin shrdlu
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### Re: almost homework, a system in 3 unknowns

PM 2Ring wrote:alessandro95, are you familiar with Cardano's & Vieta's methods for solving cubics?
For this cubic there's a more elegant proof that there's only one real solution.

Here's a question.
Spoiler:
For f continuous and differentiable on R, if f has n zeroes then what is the minimum number of zeroes f' can have?

[Edited to put question in spoiler.]

McPgr
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### Re: almost homework, a system in 3 unknowns

jaap wrote:
Spoiler:
Think about cubic polynomials.

Spoiler:
Cubics always have at least one real valued root.

Spoiler:
What are the coefficients of the cubic you get when you expand (x-r1)(x-r2)(x-r3) ?

The cubic polynomial equation will be
[math]2x^3+3x^2+6x-12=0[/math]
I failed to find any real root (not real too)

Dopefish
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### Re: almost homework, a system in 3 unknowns

There's definitely a real root. Really, just think about the graph of a cubic (with real coefficients anyway), on one side it's going off to positive infinity and the other it goes down to negative infinity, and cubics are continuous so theres definitely a crossing somewhere.

Actually finding it analytically can sometimes be tricksy, but there's always at least one real root as stated.

McPgr
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Joined: Wed Oct 20, 2010 2:28 am UTC

### Re: almost homework, a system in 3 unknowns

Dopefish wrote:There's definitely a real root.

Sure. There's one positive root (one change of sign) that must be real since complex roots appear in pairs. But it's not rational and some luck is needed to find it

letterX
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### Re: almost homework, a system in 3 unknowns

McPgr wrote:But it's not rational and some luck is needed to find it

Or, like, use the cubic formula. Or find it numerically by, e.g., Newton's method. Not sure what "luck" has to do with it, unless you're picking real numbers randomly and seeing if they're roots.

If so, "good luck" with that?