## Mathematical solution to two quadratics

chemoleo - 22-4-2008 at 18:18

I'm trying to find the solution to a quadratic equation (ax^2 + bx + c) whose intrinsic terms b and c contain terms that are in turn only solvable by a quadratic solution.

I'm not a mathematician (have mercy), and I'm happy to have arrived at this point by my own so far.
How do I solve the below for [PL] ?

Any help welcome. Please do not be cryptic!
Spent enough time on this already.
Thanks.

roamingnome - 22-4-2008 at 19:28

MAPLE mathcad or one of the newer TI calculators have a "solve for" variable function.

this may be considered cheating or the wussy way out, but it does get the job done if the solutions are real....

they may even spit out imaginary complex solutions these days...

the TI that i saw has a step wise solver that gave you each step so it seemed like you were learning.. not just getting answers...

chemoleo - 22-4-2008 at 19:39

Well, I'm lead to believe that a cubic equation can be formed of eqn 3, but I've no idea how, and how to form the solution of such a cubic eqn.

I need this very solution for curve fitting of real data, so I couldnt' use any shortcut or shortcut protocol...

I know it can be done but I don't know how.

12AX7 - 23-4-2008 at 00:11

So [PC] == x_[PC]?

Try factoring the quadratic (in x) to seperate the factor of [PC]. Remember that the quadratic equation can be undone, quite easily, by reversing some algebra and un-completing the square.

Would you provide the original equations?

Tim

len1 - 23-4-2008 at 02:33

Equations of type

y^2 + a x y + b y^2 + c x + d y + e = 0

are known as conics. Performing a translation

y' = y + alpha
x' = x + beta

we can choose these so c and d = 0

A rotation on (x, y) then removes the xy term, so the equation reduces to one of three cases

y^2 + a' x^2 = b' an ellipse

y^2 - a' x^2 = b' a hyperbola

y^2 + a' x = 0 a parabola. (or x <-> y)

The first equation represents a rotated and translated conic.

If we have two of these to solve simultaneously, we are looking for the intersection of two conics, which can have 0, 2 or 4 real solutions. One of these can be centred by transformations to one of the above forms.

The second will stay of the general form

y^2 + a'' x^2 + b''y x + c'' x + d''y + e''=0

where the parameters are linear combinations of your a,b,s,t, which are easy enough to get.

Unfortunately its pointless me calculating it for you as you can easily see by substituting the reduced form above, say

y = sqrt(b' - a' x^2)

into the general form, which then gives you an equation in x only, that this equation is at best a quartic (as indicated by the 4 possible solutions), but at worst of order 8 - which can be reduced to a quartic.

Unless we have a special case, the analytic form for quartics is horrendous and will tell us nothing - I cant see why you'd need it. Best to do it graphically or numerically - not analyically

[Edited on 23-4-2008 by len1]

chemoleo - 23-4-2008 at 04:28

Don't have time right now to look at this right now, but I did write down the derivations, if anyone's interested how I got there.
Thanks.

PS Len, I think the first eqn is of a different type than I represent.
But have a look at the attachment and you'll see what I mean.

Attachment: comp.pdf (31kB)

len1 - 23-4-2008 at 04:59

No, everything is right.

You can transform your equation by getting all terms not involving the square root onto one side, the other will be of the form

sqrt() + sqrt()

take the square of both sides, that eliminates the roots, but you get a quartic as I said

chemoleo - 23-4-2008 at 16:16

Ok I'll look into solving it into something simpler.
A true analytical solution would be ideal because I can then plug this into sigmaplot/origin, and just do a (nonlinear) curvefit to Fobs versus varying [Co]

How do I best go about getting a numerical solution?
Mathematica does it but I've never used it.

And I still have no idea how to integrate multiple numerical solutions for 50 datapoints into a curvefit which is fit by non-linear regression...
After all I'm only after [PL] because I can insert this into the equation describing the total observed fluorescence (see attachment).
And the total observed fluorescence is a function of Kd[PC], which is what I'm trying to determine here, so Kd[PC] has to be iterated in turn .

A zero point solution of [PL] in the range of 0 --> Po ([PL] can never be greater than Po) is determined by the observed data Fobs (to which I later make the curvefit for multiple observed Fobs datapoints), the known constants Po, Lo, Co and Kd[PC], and the unknown constant Kd[PC]...

Horrendously complicated.

Anyway, are there good easy to use packages where I can plug in about any equation, a range of datapoints (varying Co), which then spits out the numerical solution?

len1 - 23-4-2008 at 23:03

OK Ill do it in mathematica within a day or two and will send you results.
chemrox - 24-4-2008 at 21:33

trig substitution?
len1 - 25-4-2008 at 00:27

Turns out mathematica is not needed

Rewriting your equations 9 and 10

letting

kpdl = a po=b L0=c c0=d kpdc=e

Pl=x
Pc=y

9) ax = (b-x-y)(c-x)
10) ey = (b-x-y)(d-y)

so

ax (d-y) = ey (c-x)

thus

substitite this value of x into 10

ey = (b - eyc/( ad+(e-a)y ) - y) (d-y)

now multiply both sides by (ad + (e-a)y)

ey(ad + (e-a)y) = ( (b -y)(ad + (e-a)y) - eyc ) (d-y)

Ill leave you to apply the distributive rule and gather terms in y^3, y^2 etc. As you see we have a cubic - so its one better than a quartic.

The solution to a cubic

y^3 + my^2 + ny + p = 0 is given in wikipedia - I cant write it out in this editor. Substitute the values of m, n, p from equation above

et voila!

JohnWW - 25-4-2008 at 15:27

Chapter 2 of Perry's Chemical Engineers' Handbook (see the Chemical Engineering Books section in the References forum for links for downloading the 7th and 8th editions) has the methods for algebraic solution of cubic and quartic equations. There is also a trigonometric/hyperbolic function solution of cubic equations, but it is not included.
sparkgap - 29-4-2008 at 01:52

@len1 and everybody else doing mathematical stuff:

Have a look at the LaTeX equation editor: just save the image generated after converting the LaTeX input and upload it to the "scipics" account.

@JohnWW: the trigonometric form for solving the cubic equation is useful only for the "all real roots" case (a.k.a. casus irreduciblis).

@chemoleo: If you're still having a hard time making heads or tails of this, I'll see if I can write up something readable.

sparky (~_~)

chemoleo - 29-4-2008 at 17:53

Thanks everyone for the replies.
I won't have time to look into this properly until next week, but then I'll need that very solution for my experiments! Let's hope it is all just like you say, that there is truly an analytical solution!

JohnWW - 3-5-2008 at 04:28

 Quote: Originally posted by sparkgap @JohnWW: the trigonometric form for solving the cubic equation is useful only for the "all real roots" case (a.k.a. casus irreduciblis).

For the case of one real root and two complex roots, the hyperbolic equivalent of the trigonometric solution of cubic equations can be used, as an alternative to the algebraic solution.

JohnWW - 6-5-2008 at 15:17

Thanks. BTW I quite often have to solve cubic and quartic equations in some areas of my work, algebraically or by trigonometric/hyperbolic substitution, e.g. for deriving forecast internal rates of return (IRR) of business projects over 3 or 4 years; but not, in general, for greater numbers of years. I also sometimes have to solve cubic equations in order to solve the Van-Der-Waals Equation of state of "non-ideal" gases. See:
http://en.wikipedia.org/wiki/Cubic_equation
http://en.wikipedia.org/wiki/Quartic_equation
http://mathworld.wolfram.com/CubicEquation.html
http://mathworld.wolfram.com/CubicFormula.html
http://www.sosmath.com/algebra/factor/fac11/fac11.html
http://home.scarlet.be/~ping1339/cubic.htm
http://eqworld.ipmnet.ru/en/solutions/ae/ae0103.pdf
http://mathforum.org/dr.math/faq/faq.cubic.equations.html
http://www.britannica.com/eb/topic-145710/cubic-equation
http://www.glue.umd.edu/~nsw/ench250/cubiceq.htm

The resolvent equation of a quartic is a cubic, with coefficients in terms of those of the quartic. However, the above references are very short on discussing the solutions of cubics using trigonometric or hyperbolic functions, which make use of the triple-angle formula.

For all practical purposes, higher-degree equations (than cubic and quartic), e.g. quintic, can be solved only numerically or graphically, e.g. by Newton's method. It was proven, by Galois, Abel, and Ruffini, that general quintic (5th-degree) equations cannot be solved algebraically, because the resolvent equation, for the coefficients of what should be a lower-degree equation, is 6th-degree instead; and similarly with higher-degree equations. They can theoretically be solved by use of elliptic functions, which are the doubly-periodic homolog of trigometric and hyperbolic functions, as discovered by Hermite and Tchirnhausen (1858), but it is difficult. Palindromic quintics can be algebraically reduced to a lower degree; and a special class of quintics with only odd powers of the variable can be solved by trigonometric or hyperbolic substitution. See:
http://en.wikipedia.org/wiki/Quintic_equation
http://mathworld.wolfram.com/QuinticEquation.html
http://forums.wolfram.com/student-support/topics/4901
http://www.bookrags.com/research/quintic-equations-wom/
http://www.bookrags.com/Quintic_equation
http://www.math.dartmouth.edu/~doyle/docs/icos/icos/icos.htm...
http://van-der-waals.pc.uni-koeln.de/quartic/quartic.html
http://www.euclideanspace.com/maths/algebra/equations/polyno...
http://www.geocities.com/titus_piezas/quintics.html
http://www.sciencecentral.com/site/502040
http://www.curiousmath.com/index.php?name=PNphpBB2&file=...
http://www.cs.rice.edu/~vardi/comp409/lec9.ps
http://eric.ed.gov/ERICWebPortal/recordDetail?accno=EJ769610
http://www.kerala.com/wiki-Quintic_equation
http://www2.math-4-u.com/child/daisu.html
http://www.csh.rit.edu/~topher/math/quinticgallery.html (list of special solvable quintics)
http://www.encyclopedia.com/doc/1O82-quinticequation.html
http://econpapers.repec.org/article/blamanch2/v_3A42_3Ay_3A1...
http://www.math.rochester.edu/u/faculty/doug/oldcourses/237f...
http://projecteuclid.org/handle/euclid.bams/1183486686
http://bbs.sachina.pku.edu.cn/Stat/Math_World/math/q/q111.ht...
http://www.learner.org/channel/courses/mathilluminated/units...
http://www-history.mcs.st-and.ac.uk/history/Printonly/Abel.h...
http://digital.lib.lehigh.edu/cdm4/remain_viewer.php?DMTHUMB...
http://sci.tech-archive.net/Archive/sci.math/2006-06/msg0000...
http://etc.usf.edu/lit2go/contents/2800/2887/2887_txt.html
http://www.blackwell-synergy.com/doi/abs/10.1111/j.1467-9957... (abstract)
http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?ha...

[Edited on 8-5-08 by JohnWW]

len1 - 7-5-2008 at 15:03

Quintic and higher cant be solved by root extractions. However the resolvent being of higher degree than original equation is I dont believe a proof. It just means the resolvent method doesnt work.

The proof Galois gave, although he didnt say this, I have boiled down to one simple fact

The symmetries admitted by roots of all quintics and higher (Galois group) is greater than the symmetry of rational fields adjoined by extraction of 5th and higher roots,

Geomancer - 7-5-2008 at 15:59

Careful there, len1. The (Galois) symmetry group associated to most quintics is qualitatively different than that of lower order polynomials. It is possible to construct polynomials, solvable by radicals, with very large and somewhat complex Galois groups.

[Edited to clarify the last sentence]

[Edited on 8-5-2008 by Geomancer]

len1 - 7-5-2008 at 20:25

OK. So I have missed the word 'generally'. But we all know that x^5 - 1 = 0 for example is solvable in radicals. So I think this is just pedantics. You need not specify any more than is necessary for whoever you are communicating with to understand you. Everything we say is precisely thus. Especially true for forum banter. Like Feynman said, theres no science in pedantics, only pseuoscience.

[Edited on 8-5-2008 by len1]

Geomancer - 8-5-2008 at 06:08

No need to take offense. My point was that the term "greater than" is difficult to interpret. The size of the group is irrelevant; it's the structure that counts.
len1 - 8-5-2008 at 07:04

Did I say the size of a group anywhere? These are your words.

I pointed out several times that if one has to constantly look over the shoulder when adding meaningful posts to a forum such as this, from people who want to nit-pick but dont actually add anything useful themselves, then one hardly need bother. I have no problem with valid criticism, in fact there would be no point in posting otherwise. But youre in fact engaging in a game which a few here play, build yourself a straw charicature and shoot it down for personal gratification.

If someone doesnt know what I meant by greater than, and wants to ask politely, then I have no problem in engaging in dialogue with such people. If having obtained that explanation one thinks its wrong for a reason he can demonstrate - I have no problem with that.

[Edited on 8-5-2008 by len1]

Geomancer - 8-5-2008 at 10:12

*Sigh* A personal attack? Really? I stand by the quality of my posts, both in general and in regards to the posts in this thread.
 Quote: Originally posted by len1 Did I say the size of a group anywhere? These are your words.

Yes, because your words were meaningless. I'm familiar with the criterion of the solvability of the Galois group, and, while I assume that's what you refer to, I can't easily relate that to what you wrote. Use of "greater than" implies a [partial] order. While I can think of several such orders for groups (and more for field extensions), mostly related in some way to size (order, composition length, inclusion as an isomorphic subgroup, etc.), I don't know of any natural one that implies solvability. Certainly it's not standard nomenclature. I posted to clarify what I found unclear, thus increasing the meaning in your post.

 Quote: Originally posted by len1 I pointed out several times that if one has to constantly look over the shoulder when adding meaningful posts to a forum such as this, from people who want to nit-pick but dont actually add anything useful themselves, then one hardly need bother. I have no problem with valid criticism, in fact there would be no point in posting otherwise. But youre in fact engaging in a game which a few here play, build yourself a straw charicature and shoot it down for personal gratification.

Perhaps you should consider polishing that particular mirror a bit less. It seems to be confusing you.
 Quote: If someone doesnt know what I meant by greater than, and wants to ask politely, then I have no problem in engaging in dialogue with such people. If having obtained that explanation one thinks its wrong for a reason he can demonstrate - I have no problem with that.

Fair enough, what did you mean? As to politeness, I fail to see what you considered impolite in my original post. Was it that I referred to you by name? That was meant to give credit for introducing the concept in the discussion.

Lest I be accused of pursuing a personal argument without proportional contribution in a public thread, below is a brief description of what Galois theory is all about. It's a bit of a work in progress, so anyone may feel free to request clarifications or corrections. Normally, of course, this material takes a long time to set out. The purpose here is simply to allow the curious to get a taste of the issues and techniques involved. Needless to say, I've omitted the proofs.

The Tornadic Tour of Galois Theory
The main objects studied in Galois Theory are not equations, or even polynomials, but rather fields, specifically finite field extensions. Informally, a field is an algebraic structure where one may add, subtract, multiply and divide (except by 0, of course) normally. Familiar examples are the rational numbers, the real numbers, and the complex numbers.

Consider a polynomial P in one variable with rational coefficients. Further assume P cannot be factored as a product of smaller polynomials with rational coefficients. Clearly P cannot factor into linear components ("split") with rational coefficients. There are, however, fields (bigger than the rationals) where it does split. In fact, we may construct such. Formally adjoin a symbol ("w") to the rational field, and let w satisfy (formally) P(w)=0. It turns out that the object so created will itself be a field, the so-called extension of the rationals by w. Since P(x)=0, (x-w) divides P, so P/(x-w) is a new polynomial, and we may iterate the process to obtain a (minimal) field where P splits ("splitting field of P"). It is important to note that different polynomials may have the same splitting field. For example x^2-2 and x^2+2x-1.
From the construction one may guess at the fact that all roots of a polynomial are equivalent with respect to the rational field. Mathematically, one may produce a function h ("field automorphism fixing the rationals") from the splitting field to itself, satisfying h(a+b)=h(a)+h(b), h(ab)=h(a)h(b), h(x)=x for all rational numbers x, such that given a pair (x1, x2) of roots of P h(x1)=x2. It is easy to see that any field automorphism fixing the rationals of a splitting field of P must permute the roots of P. Not all permutations must correspond to such automorphisms, though. For example, x^3-3x-1 has a splitting field with exactly 3 automorphisms fixing the rationals (i.e. only half the permutations of the roots generate automorphisms). The set of all automorphisms of a field fixing the base field (the rationals, for the sake of this discussion) is called its "Galois Group" (technically, one usually looks at this as an abstract group. That is beyond the scope of this discussion).

The fundamental insight regarding the solvability of polynomials is that when one extends a field by a root of x^n-a, the Galois group of the field changes by what are known as "cyclic extensions", and so the Galois groups of the splitting fields of polynomials must be obtainable by a sequence of such extensions. But not all finite groups are so obtainable, so polynomials with such splitting fields are not solvable.

[Edited on 8-5-2008 by Geomancer]

JohnWW - 8-5-2008 at 13:53

 Quote: Originally posted by len1 OK. So I have missed the word 'generally'. But we all know that x^5 - 1 = 0 for example is solvable in radicals.

That (which is effectively a palindromic quintic, the roots being the five 1/5th roots of 1) factorizes to
x^5 - 1 = (x - 1)(x^4 + x³ + x² + x + 1) = 0
so that the roots are 1 and the four complex roots of an easily solvable palindromic quartic. By dividing the quartic through by x², it becomes (x + 1/x)² + (x + 1/x) - 1 = 0, which is a quadratic in (x + 1/x), solving which gives another quadratic:
x² - ½(-1 ± sqrt(5))x + 1 = 0
the four solutions of which are the complex fifth roots of 1:
x = ¼[-1 ± sqrt(5) ± isqrt(2(5 ± sqrt(5))]
where i = sqrt(-1). The coefficients of the real and complex parts are the sin and cos of ¶/5 and 2¶/5 radians.

[Edited on 9-5-08 by JohnWW]

len1 - 8-5-2008 at 14:03

Ordinarily I dont reply to people obviously out to beat their chests and pick a fight with provocations such as these

Quote:

Originally posted by len1
Did I say the size of a group anywhere? These are your words.
 Quote: Yes, because your words were meaningless.

Ill let people judge whether my comments regarding building oneself a straw man to shoot down for personal gratification are justified on the basis of that question and answer.

 Quote: Fair enough, what did you mean?

Shouldnt that question have come first rather than several posts into your argument?

However the substance you wrote does I believe contain some merit in condensing the essence of Galois aims (I assume these are your words and not taken from somewhere), so Ill let this be an exception and post your very own answer for the sake of others who maybe are still following this thread

 Quote: so the Galois groups of the splitting fields of polynomials must be obtainable by a sequence of such extensions. But not all finite groups are so obtainable, so polynomials with such splitting fields are not solvable.
So the number of finite groups corresponding to polyonomials is greater than (or more precisely is a superset of) the number of groups obtainable by such extensions.

Im sure a person with your 'talents' will now find a new way to attack this statement. Im not interested in this. I have obviated your motives but I will not try to find pedantic holes in your exposition of the subject. You do say it normally takes a long time to set out to try and defend yourself against such. I just note you had the benefit of three paragraphs (which most people here wont be able to follow) as opposed to my one sentence - and still are worried it might leave you open to 'misunderstandings'.

[Edited on 8-5-2008 by len1]

Geomancer - 8-5-2008 at 16:44

@len1: You impugned my integrity multiple times over the course of two posts before I chose to write with anything other than complete politeness. You continue to do so, despite being told numerous times, on numerous occasions, that no offense was intended. You accuse me of plagiarism, building straw-men, idle pedantry and pseudoscience, yet somehow claim that I'm out to pick a fight? I stand by what I wrote and will vigorously defend my integrity against all such unprovoked assaults.

If I wrote something incorrect, please do point it out, as this only serves to increase the value of the forum's content.

JohnWW - 16-5-2008 at 00:48

### Solution?

chemoleo - 19-5-2008 at 16:46

Ok, I've done some work on this:
With the hint from len1 I did the following:
(I'd appreciate someone checking, I'm utterly paranoid I made a mistake somewhere...pleeeease! )
See attachment.

Substituting this into the cubic solution is ok, but now I have another problem: I need to find some regression software that can handle the third root! I couldn't find such a function in Sigmaplot - is anyone aware of graphing software that does advanced regression offering a third root function?

Please be aware that I fit real data (as Fobs) as in

Fobs = Fmin(Lo-[PL]) + Fmax([PL])

where Fobs are about 50 data points.
I therefore have to plug in the cubic solution into each [PL] in the eqn above, and by varying the KdPL (substituted as a) I get the best fitting curve (and the KdPL therefore is value I'm after)... Not sure if this is all making sense, but I did explain it in one of the attachments above.

Anyway, I'm now stuck with trying to find regression software that offers the third root... anyone?

Thanks!