# System of equations difficulty

1. Apr 27, 2006

### tehno

$$-44x^2 -2y^2 - 47xy + 45x + 2y + z =0 ; -2x^2 + y^2 + 43 xy + 2x + z =0; x^2 - 44y^2 - 41xy - 2x + 43y + z =-1$$
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:uhh:
Can we obtain all REAL solutions of this system without using numerical math tools?
I have problems with it and would appreciate help ,advice or your solutions.

Thanks,
tehno

Last edited: Apr 27, 2006
2. Apr 27, 2006

### tehno

$$-44x^2 - 2y^2 - 47xy+45x+2y+z=0;$$
$$-2x^2 + y^2 + 43xy + 2x + z =0;$$
$$x^2 -44y^2 - 41xy -2x +43y + z=-1$$
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3. Apr 27, 2006

### Curious3141

Wow, that's an ugly problem. Thank goodness the z term is linear and not mixed in with the other terms, that makes things doable.

There is of course an analytical way to do it, but it's pretty brute force. Let's say the equations are (1), (2) and (3) from top to bottom the way you wrote them. The first order of business is to get rid of the z term by taking (1) - (2) and (2) - (3) to give two new equations in x and y only. Let's call those (A) and (B)

It's late and I'm very sleepy so I could be wrong, but I don't think (A) and (B) are going to factor nicely. So just force it with the quadratic formula. Treat (A) and (B) like quadratic equations in x and treat y as if it's part of the coefficients/constants. By doing this you can directly solve for x in terms of y in both (A) and (B). Let's call the rearranged equations (C) and (D).

Now you can equate (C) and (D), and you finally have a single equation in terms of one variable (y). Remember that when you solved the individual quadratics to get x in terms of y, there should be plus/minus (+/-) terms , which means you actually have to consider four separate permutations in equating (C) and (D) separately. You have no choice but to do them one at a time.

In each case, I think you can manipulate them to become quadratics in y. Remember, there's squaring involved, with the introduction of possible redundant roots, so you need to check for those.

After getting all the possible values for y in each of the 4 cases and the corresponding values for x, put those back into the equation and get z.

It should be a fairly horrid exercise on the whole. Sure you want to try this ?

4. Apr 28, 2006

### Curious3141

It's worse than I initially realised. You'll have to solve a quartic with *large* coefficients.

For one of the cases (taking the plus signs in each x = f(y) expression), you get :

$$7596y + 78 = 6\sqrt{7596y^2 - 7404y + 1849} - 84\sqrt{7596y^2 - 156y + 4}$$

The only analytical way to solve that is to square both sides once, leave the big square root term on one side while moving everything to the other side and squaring again. The y^4 term does not cancel out.

The good news is that you can probably dump the quartic into Mathematica or Maple and get surd answers for y. Just remember to eliminate the redundant roots from the repeated squaring.

You'll need to do that for each of 4 cases (including the case above). Then working out z becomes an easy matter.

If you're brave enough to want to try solving this all by hand, here's something you'll need : http://www.karlscalculus.org/quartic.html

5. Apr 29, 2006

### tehno

Thanks for your input and help Curious3141!
The Quatric order is the last resort to do it by "hand" in general case of algebraic equation (according to Abel-Galois theory ,its solutions still can be "expressed" finitely in radicals).
Usually ,the Ferraris method of factorisation in two quadratic polynomials is recommended.
As of solutions,I managed to obtain root real solutions (double?) as:
x=1/3
y=1/3
z=-16/3

However I must ask you again :Are there more real solutions I forsaw somehow?
Can you cross check that for me?

Last edited: Apr 29, 2006
6. Apr 29, 2006

### Curious3141

That's very good that you managed to find the "simple" real roots like that - did you do that by inspection alone ? If so, that's really good, because I didn't see those roots.

y = 1/3 does not satisfy my equation in y, my y is going to be irrational. :grumpy: My equation might be wrong, but since I don't have the scrap paper anymore, I am unable to check where I went wrong. It's also possible my equation is correct and there are irrational roots to the equation.

In all likelihood, there are other "nice" solutions apart from the solution you found. The easiest way to look for all the solutions is to put this into Mathematica.

7. Apr 30, 2006

### tehno

Well,I don't have that software.
That's why I asked for help in a cross check in first place ))

But I have an interesting question of even higher importance:

Suppose we have to charaterize set of all REAL solutions for system of two equations:

$$f(x,y)=0$$
$$g(x,y)=0$$
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If $$f(x,y),g(x,y)$$ are both polynomials of order n>4 such
that $$f(x,y)=g(y,x)$$ is satisfied,and the polynomials have rational coeficients,
what theorem(s) can answer the question concerning the real solutions set characterisation and number of real solutions of the system judging just on the base of the coeficients and their signs?

Edit:{Of course,n is also known }

Last edited: Apr 30, 2006
8. Apr 30, 2006

### Curious3141

Ah, but you see, I don't have the software either, otherwise I would certainly have been able to help you out.

Maybe someone else will step in here ?

I'm afraid that's a little out of my depth. But if the degree of the polynomial is greater than 4 (meaning 5 and above), then there is no guarantee an analytical solution exists. I believe there are classes of quintics (and higher orders) where such solutions can be found, but I am not well up on that theory. I assume the same considerations should apply to simultaneous equations of such high order (except that when one "merges" the equations so that they get a single polynomial in one variable, the degreee is likely to be higher than either of the two separate equations).

I'm afraid I cannot really help with the question of the simultaneous equations in arbitrary order - maybe one of the mathematicians here can assist you better.

Considering only single variable polynomial equations of degree 4 and below (quartics and simpler), a guaranteed analytical solution exists, even though it's tedious to find. I just put up a tutorial linking a good site giving the general method of solution, check out this reference : https://www.physicsforums.com/showthread.php?t=119284

If you don't want to solve, but only want to ascertain how many roots are real or complex, that's fairly simple. It becomes a simple problem of reduction and deduction.

Let's say we have a quartic in x called f(x). For simplicity, always take the coefficient of x^4 to be positive (in fact, let it be positive one).

f(x) = 0 can have no real solutions (2 sets of complex conjugate roots), 2 real and 2 complex solutions or 4 real solutions. (Note that repeated roots are possible). The problem is to determine which case holds.

The easiest way is to sketch the curve and consider the turning points of the curve. The quartic is guaranteed to either one or three turning points.

If the quartic has only one turning point, then the turning point can occur above (that is, f(x) is positive at that point), at (tangential to, meaning f(x) is zero at that point) or below the x-axis (f(x) negative at that point). If it occurs above, there are no real roots (hence 4 complex roots). If it occurs below, then there are exactly 2 distinct real roots (and 2 complex roots). If it occurs tangential to the x-axis, then there are repeated real roots.

If the quartic has three turning points and they all occur above the x-axis, there are no real solutions. If all 3 turning points occur below the x-axis, there are 2 distinct real roots. And if two turning points occur below the x-axis and one occurs at or above the x-axis, there are 4 real roots (in the case where the turning point forms a tangent to the axis, there is a repeated root). Simple, no ?

But how to find the turning point(s) ? For this, examine the derivative of the quartic.

At the turning points, f'(x) = 0. The problem is reduced to solving a cubic. The cubic is guaranteed to have one real root. In addition it may have another 2 real roots or a pair of complex conjugate roots.

You can either simply solve the cubic, or repeat the process to deduce how many real roots the cubic has. A cubic function has either no turning points or two turning points. It may also have a single point of inflexion. You can determine which case holds by examining the derivative of the cubic. f''(x) is a quadratic function. If the discriminant of the quadratic is negative, then there are no stationary points (turning or inflexion points) for the cubic f'(x). If the discriminant is zero, then there is a single inflexion point. If the disc. is positive, there are 2 real turning points on the cubic. For a cubic of the form $$x^3 + bx^2 + cx + d = 0$$, the problem reduces to determining the sign of $$b^2 - 3c$$.

You can go through the same logical process to determine, based on the position of the turning points, whether the cubic has a single real root or 3 real roots. Then use those deductions to work out the nature of the roots of the quartic.

This process (of determing merely the nature of the roots) should be generalisable to any degree of polynomial equation. The difficulty lies in actually finding those roots.

Last edited: Apr 30, 2006
9. May 7, 2006

### tehno

Hey,what are you waiting guys!?
Nobody with Mathematica here?

10. May 9, 2006

### tehno

I really don't want go and get pyrate softwares..
Is there any shareware available on the web for a download that could help me for the task ?

11. May 9, 2006

### Lee

[[x = 1/3, y = 1/3, z = -16/3], [x = 1/3*RootOf(44521*_Z^3-133563*_Z^2-8862*_Z+29,label = _L1), y = 211/45*RootOf(44521*_Z^3-133563*_Z^2-8862*_Z+29,label = _L1)^2+1/45-647/45*RootOf(44521*_Z^3-133563*_Z^2-8862*_Z+29,label = _L1), z = -1/633]]

Says maple.

12. May 9, 2006

### Curious3141

The "Root of..." expressions, are they referring to the roots of another cubic ? Is it possible to get Maple to give explicit solutions instead ?

13. May 10, 2006

### J77

The rootsof bit has to do with complex solutions, iirc...

There is also a command to give more explicit solutions, ie. expand the rootsof...

But it's a long time since I've used Maple...

14. May 10, 2006

### dextercioby

I'd suggest a graphical approach to the subject.

(see post #3 for reference). I'd put the resulting (A) and (B) equations in a canonical form of ellipse (possible circle) and do the drawing and see whether the 2 closed curves intersect. If so, i'd get approximate values for intersection points's coordinates => approximate sollutions to the initial system of equations.

Daniel.

15. May 10, 2006

### J77

Graphically

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16. May 10, 2006

### Nimz

If you have f(x,y)=g(y,x), then your system of equations is f(x,y)=0 and f(y,x)=0. If there are solutions that are not of the form y=x, there are related solutions obtained by switching x and y. I see no guarantee that you will always get "nice" real solutions (that is, solutions that can be written with radicals) when you do have real solutions. Then again, I'm not familiar with the conditions for nice solutions to quintic and higher degree functions, so I may be missing something.

17. May 12, 2006

### tehno

Thanks everybody,
Especially to Lee ,who confirmed by Maple that the only real solutions are x=y=1/3,z=-16/3.

To Nimz:OK,but I was thinking of some powerful theorems .Say Rieman-Roch formulae from algebraic geometry that could possibly answer questions relating number and character of set of real solutions,without trying to solve polynomials!.
Maybe there's is something similar for systems I'm intersted in.