# Is there a more general quadratic formula?

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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?

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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.

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That equation does have constants; they are all 1.

mm nope, sorry, and 0=/=1+1+1

Edited by SFNQuestions
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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 $ax^2 + bx + c$ ? 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 by wtf
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How is that different than $ax^2 + bx + c$ ? 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 by SFNQuestions
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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.
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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 by SFNQuestions
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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.

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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 ax2+ bx + c = 0, a method that won;t work with multiple variables, unless of course you have multiple unique equations).

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Right OldChemE. +1

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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 by SFNQuestions
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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

Ax2 + Bx + Cxy+ Dy2 + Ey + F = 0

You should write your expressions in this form for as many variables as you have.

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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

Ax2 + Bx + Cxy+ Dy2 + 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 by SFNQuestions
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I see no point discussing further, considering your response to myself and others when we try to help.

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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 by SFNQuestions
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• 4 months later...

The 'quadratic formula' you reference initially says that the solutions to $ax^2+ bx+ c= 0$ are given by $\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$. There is ​no ​requirement that a, b, and c be constants, just independent of x. The solutions to $0=xz^2+yz+w$ are given by $z= \frac{-y\pm\sqrt{y^2- 4xw}}{2x}$.

Edited by Country Boy

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