SFNQuestions 17 Posted January 27 I want to use a quadratic equation that uses not constants, but variables, like 0=xz^2+yz+w. Does the regular formula work for that? Share this post Link to post Share on other sites

Acme 496 Posted January 27 I want to use a quadratic equation that uses not constants, but variables, like 0=xz^2+yz+w. Does the regular formula work for that? That equation does have constants; they are all 1. Share this post Link to post Share on other sites

SFNQuestions 17 Posted January 27 (edited) That equation does have constants; they are all 1. mm nope, sorry, and 0=/=1+1+1 Edited January 27 by SFNQuestions Share this post Link to post Share on other sites

wtf 66 Posted January 27 (edited) I want to use a quadratic equation that uses not constants, but variables, like 0=xz^2+yz+w. Does the regular formula work for that? How is that different than [math]ax^2 + bx + c[/math] ? In that expression, the coefficients are constant for a given polynomial, but as they vary they generate the space of all quadratic polynomials. How do you interpret your equation differently? Edited January 27 by wtf Share this post Link to post Share on other sites

SFNQuestions 17 Posted January 27 (edited) How is that different than [math]ax^2 + bx + c[/math] ? In that expression, the coefficients are constant for a given polynomial, but as they vary they generate the space of all quadratic polynomials. How do you interpret your equation differently? Well, they aren't necessarily constant with respect to the variable you're solving for, you can't always just pretend variables as constants, so it may be okay to always use the quadratic formula in that manner, or there may be exceptions in which case there is no garuntee the formula works. If the intermediate algebra allows for the separation of variables, then it would be valid, but if there is no known method of separating the variables but still pretend the variables are constants to loophole the problem, you will find your solution is completely invalid. Edited January 27 by SFNQuestions Share this post Link to post Share on other sites

Acme 496 Posted January 27 That equation does have constants; they are all 1.My bad; I meant to say the coefficients are all 1. Well, they aren't necessarily constant with respect to the variable you're solving for, you can't always just pretend variables as constants, so it may be okay to always use the quadratic formula in that manner, or there may be exceptions in which case there is no garuntee the formula works. If the intermediate algebra allows for the separation of variables, then it would be valid, but if there is no known method of separating the variables but still pretend the variables are constants to loophole the problem, you will find your solution is completely invalid.Variables -by definition- can vary in their values. Constants -by definition- do not vary in their value. One cannot 'just pretend variables as constants'. Oh... but you're loop holing it, so it makes perfect sense. Share this post Link to post Share on other sites

SFNQuestions 17 Posted January 27 (edited) My bad; I meant to say the coefficients are all 1. Oh... but you're loop holing it, so it makes perfect sense. Which is what I said you can't do...so you should be agreeing with me...Do you understand English? Are you using some outdated online translator? Edited January 27 by SFNQuestions Share this post Link to post Share on other sites

Sriman Dutta 44 Posted January 27 You want to use variables in place of a,b and c ? Right? That means their values are undeterminable. And such an equation will not hold true. Share this post Link to post Share on other sites

OldChemE 25 Posted January 27 In order to find a unique solution for each variable you must have as many different relationships as you have variables. The quadratic Equation works because there is only one variable (usually x, but doesn't have to be). If you have three variables in a single equation, such as you suggest, you can find an essentially infinite set of values that will solve the equation, but to find the unique solution you need three different equations all using the same three variables. That means there is no shortcut equation like the quadratic equation (the Quadratic equation is simply derived by the method called "Completing the Square" starting with the ax^{2}+ bx + c = 0, a method that won;t work with multiple variables, unless of course you have multiple unique equations). 2 Share this post Link to post Share on other sites

Sriman Dutta 44 Posted January 27 Right OldChemE. +1 Share this post Link to post Share on other sites

SFNQuestions 17 Posted January 27 (edited) Right that's all good stuff, but, I don't need to determine the specific values of the variables, hence why I said "not" to treat them as constants, obviously you can tell just by looking at it that the system is undetermined. My concern, as I said, is the relationship between the variables, not the specific values. But doing some more digging I've found that the method is valid in this specific case, which is perfect because I used that to show the inverse of the trigonometric functions are complex logarithms of radicals. This is more to due with u-substitution than anything else, but with complex functions you might have a z and an x and a y and u and v and so on. Edited January 27 by SFNQuestions Share this post Link to post Share on other sites

studiot 1183 Posted January 27 Is there a more general quadratic formula? Of course there is but you need to distinguish between expressions, formulae, equations, identities, variables etc. For instance the general quadratic equation in two variables is Ax^{2} + Bx + Cxy+ Dy^{2} + Ey + F = 0 You should write your expressions in this form for as many variables as you have. Share this post Link to post Share on other sites

SFNQuestions 17 Posted January 27 (edited) Of course there is but you need to distinguish between expressions, formulae, equations, identities, variables etc. For instance the general quadratic equation in two variables is Ax^{2} + Bx + Cxy+ Dy^{2} + Ey + F = 0 You should write your expressions in this form for as many variables as you have. The way I wrote it is as many variables as I have. But, the more general the merrier I suppose, I don't have anything against a more general expression. Edited January 27 by SFNQuestions 1 Share this post Link to post Share on other sites

studiot 1183 Posted January 27 I see no point discussing further, considering your response to myself and others when we try to help. Share this post Link to post Share on other sites

SFNQuestions 17 Posted January 27 (edited) I see no point discussing further, considering your response to myself and others when we try to help. I also see no point in you discussing it further, seeing as how it is no one else's fault besides your own that you made careless assumptions rather than actually trying to help by simply being more cautious in analyzing the problem. Edited January 27 by SFNQuestions 3 Share this post Link to post Share on other sites

HallsofIvy 52 Posted June 23 (edited) The 'quadratic formula' you reference initially says that the solutions to [math]ax^2+ bx+ c= 0[/math] are given by [math]\frac{-b\pm\sqrt{b^2- 4ac}}{2a}[/math]. There is no requirement that a, b, and c be constants, just independent of x. The solutions to [math]0=xz^2+yz+w[/math] are given by [math]z= \frac{-y\pm\sqrt{y^2- 4xw}}{2x}[/math]. Edited June 23 by HallsofIvy Share this post Link to post Share on other sites