SFNQuestions

Is there a more general quadratic formula?

Recommended Posts

Acme    496

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
wtf    63

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 by wtf

Share this post


Link to post
Share on other sites
SFNQuestions    17

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 by SFNQuestions

Share this post


Link to post
Share on other sites
Acme    496

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. :rolleyes:

Share this post


Link to post
Share on other sites
SFNQuestions    17

My bad; I meant to say the coefficients are all 1.

Oh... but you're loop holing it, so it makes perfect sense. :rolleyes:

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

Share this post


Link to post
Share on other sites
Sriman Dutta    44

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    23

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).

  • Upvote 2

Share this post


Link to post
Share on other sites
SFNQuestions    17

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

Share this post


Link to post
Share on other sites
studiot    1131

 

 

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.

Share this post


Link to post
Share on other sites
SFNQuestions    17

 

 

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
  • Downvote 1

Share this post


Link to post
Share on other sites
studiot    1131

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

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
  • Downvote 3

Share this post


Link to post
Share on other sites
HallsofIvy    41

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 by HallsofIvy

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now