How to solve for four variables when two have first and second degrees

[JazzyJones]

Homework Statement

does anyone know how to solve this equation for just x?

x + x^2 = a^2*b^2 - 2.01j - j^2 - 15.132

it's possible right? because I'm trying to solve for four variables using four equations but two of my variables will have a first plus a second degree

The Attempt at a Solution

my original equation was in the form [((x-h)^2)/a^2] + [((y-k)^2)/b^2] = 1

i am trying to solve for h, k , a, b

 

[Dick]

Homework Statement

does anyone know how to solve this equation for just x?

x + x^2 = a^2*b^2 - 2.01j - j^2 - 15.132

it's possible right? because I'm trying to solve for four variables using four equations but two of my variables will have a first plus a second degree

The Attempt at a Solution

my original equation was in the form [((x-h)^2)/a^2] + [((y-k)^2)/b^2] = 1

i am trying to solve for h, k , a, b

For the first equation, sure you can solve for x. Completing the square is probably the easiest way. But I don't see what the first equation has to do with the second one and you certainly can't solve the second one for the four variables h,k,a,b. You only have one equation.

 

[NasuSama]

does anyone know how to solve this equation for just x?

x + x^2 = a^2*b^2 - 2.01j - j^2 - 15.132

my original equation was in the form [((x-h)^2)/a^2] + [((y-k)^2)/b^2] = 1

i am trying to solve for h, k , a, b

Bring the right side expression to the left and then, try to complete the squares. Now, try to make that equation into the given form.

 

[JazzyJones]

For the first equation, sure you can solve for x. Completing the square is probably the easiest way. But I don't see what the first equation has to do with the second one and you certainly can't solve the second one for the four variables h,k,a,b. You only have one equation.

I have x and y values. I plug the four different coordinates into the second equation and get four equations. Then I solve using quadratic for h and k, (a and b can be solved without quadratic I believe). Then start plugging them into one equation at a time correct?

 

[Dick]

I have x and y values. I plug the four different coordinates into the second equation and get four equations. Then I solve using quadratic for h and k, (a and b can be solved without quadratic I believe). Then start plugging them into one equation at a time correct?

That should work. If you have four different values of (x,y) then you should get four different equations for h,k,a,b. You should be able to solve them in principle. Sounds pretty hard though.

 

[JazzyJones]

Sounds pretty hard though.

Yea, I'm aware of that :D

 

[Dick]

Yea, I'm aware of that :D

There's probably not even a solution for all possible selections of four values of (x,y). You aren't trying to solve what might be an easier problem the hard way are you?

 

[Vadar2012]

Would be nice to see all 4 equations. You can't just put them into a matrix form and solve a^-1*b = c?

 

[JazzyJones]

Would be nice to see all 4 equations. You can't just put them into a matrix form and solve a^-1*b = c?

Ok I got them here, http://imgur.com/2KL0g
the big K and little k are the same variable, i just wrote them like that sorry. Made a note of the missing k in the second equation and rewrote the number 7.66 for 7.66h in equation 4

I realize a matrix is used for this sort of thing sadly I don't have experience with them nor understand "solve a a^-1*b = c?"
If you could show me the steps and processes, that would be cool

 

[Vadar2012]

I'd just write a code to solve it by Newtonian or something. Doing it by hand it just too much effort and too much room for error. All that substitution is just asking for problems.

 

[JazzyJones]

I'd just write a code to solve it by Newtonian or something. Doing it by hand it just too much effort and too much room for error. All that substitution is just asking for problems.

what do you mean solve it by Newtonian??