Not sure how to rearrange this formula.
$begingroup$
I have an equation in the form,
$$
z = k_1 cdot frac{k_2 xy}{left(x^2 - y^2right)^2}
$$
where $k_1$ and $k_2$ are constant. I need to rewrite it in terms of $y$ but the closest I've come is an equation in the form,
$$
x^4 - xyleft(2xy + frac{k_1 k_2}{z}right) + y^4 = 0
$$
but I can't use substitution to solve this. From another approach I get,
$$
frac{x^2 - y^2}{sqrt{y}} = sqrt{frac{k_1k_2}{z}}
$$
so I'm rather stuck...
algebra-precalculus
asked Dec 12 '18 at 10:13
John MillerJohn Miller
1387
$endgroup$
|
show 3 more comments
$begingroup$
I have an equation in the form,
$$
z = k_1 cdot frac{k_2 xy}{left(x^2 - y^2right)^2}
$$
where $k_1$ and $k_2$ are constant. I need to rewrite it in terms of $y$ but the closest I've come is an equation in the form,
$$
x^4 - xyleft(2xy + frac{k_1 k_2}{z}right) + y^4 = 0
$$
but I can't use substitution to solve this. From another approach I get,
$$
frac{x^2 - y^2}{sqrt{y}} = sqrt{frac{k_1k_2}{z}}
$$
so I'm rather stuck...
algebra-precalculus
asked Dec 12 '18 at 10:13
John MillerJohn Miller
1387
$endgroup$
$begingroup$
Do you have any reason to believe that it is even possible to rewrite the equation in such a way?
$endgroup$
– 5xum
Dec 12 '18 at 10:15
$begingroup$
No I don't even know if you can, it's not a question from a textbook if that's what you're asking.
$endgroup$
– John Miller
Dec 12 '18 at 10:17
$begingroup$
When one solves for $y$ using computer algebra, you get a horrendously complex solution.
$endgroup$
– David G. Stork
Dec 12 '18 at 10:17
$begingroup$
Oh why is this that some equations cannot be rearranged?
$endgroup$
– John Miller
Dec 12 '18 at 10:18
$begingroup$
@JohnMiller That's, in a way, the wrong question to ask. The better question would be "why should all equations be such that they can be rearranged"?
$endgroup$
– 5xum
Dec 12 '18 at 10:19
|
show 3 more comments
$begingroup$
I have an equation in the form,
$$
z = k_1 cdot frac{k_2 xy}{left(x^2 - y^2right)^2}
$$
where $k_1$ and $k_2$ are constant. I need to rewrite it in terms of $y$ but the closest I've come is an equation in the form,
$$
x^4 - xyleft(2xy + frac{k_1 k_2}{z}right) + y^4 = 0
$$
but I can't use substitution to solve this. From another approach I get,
$$
frac{x^2 - y^2}{sqrt{y}} = sqrt{frac{k_1k_2}{z}}
$$
so I'm rather stuck...
algebra-precalculus
asked Dec 12 '18 at 10:13
John MillerJohn Miller
1387
$endgroup$
I have an equation in the form,
$$
z = k_1 cdot frac{k_2 xy}{left(x^2 - y^2right)^2}
$$
where $k_1$ and $k_2$ are constant. I need to rewrite it in terms of $y$ but the closest I've come is an equation in the form,
$$
x^4 - xyleft(2xy + frac{k_1 k_2}{z}right) + y^4 = 0
$$
but I can't use substitution to solve this. From another approach I get,
$$
frac{x^2 - y^2}{sqrt{y}} = sqrt{frac{k_1k_2}{z}}
$$
so I'm rather stuck...
algebra-precalculus
algebra-precalculus
asked Dec 12 '18 at 10:13
John MillerJohn Miller
1387
asked Dec 12 '18 at 10:13
John MillerJohn Miller
1387
edited Dec 12 '18 at 13:23
John Miller
asked Dec 12 '18 at 10:13
John MillerJohn Miller
1387
asked Dec 12 '18 at 10:13
John MillerJohn Miller
1387
asked Dec 12 '18 at 10:13
John MillerJohn Miller
1387
1387
$begingroup$
Do you have any reason to believe that it is even possible to rewrite the equation in such a way?
$endgroup$
– 5xum
Dec 12 '18 at 10:15
$begingroup$
No I don't even know if you can, it's not a question from a textbook if that's what you're asking.
$endgroup$
– John Miller
Dec 12 '18 at 10:17
$begingroup$
When one solves for $y$ using computer algebra, you get a horrendously complex solution.
$endgroup$
– David G. Stork
Dec 12 '18 at 10:17
$begingroup$
Oh why is this that some equations cannot be rearranged?
$endgroup$
– John Miller
Dec 12 '18 at 10:18
$begingroup$
@JohnMiller That's, in a way, the wrong question to ask. The better question would be "why should all equations be such that they can be rearranged"?
$endgroup$
– 5xum
Dec 12 '18 at 10:19
|
show 3 more comments
$begingroup$
Do you have any reason to believe that it is even possible to rewrite the equation in such a way?
$endgroup$
– 5xum
Dec 12 '18 at 10:15
$begingroup$
No I don't even know if you can, it's not a question from a textbook if that's what you're asking.
$endgroup$
– John Miller
Dec 12 '18 at 10:17
$begingroup$
When one solves for $y$ using computer algebra, you get a horrendously complex solution.
$endgroup$
– David G. Stork
Dec 12 '18 at 10:17
$begingroup$
Oh why is this that some equations cannot be rearranged?
$endgroup$
– John Miller
Dec 12 '18 at 10:18
$begingroup$
@JohnMiller That's, in a way, the wrong question to ask. The better question would be "why should all equations be such that they can be rearranged"?
$endgroup$
– 5xum
Dec 12 '18 at 10:19
$begingroup$
Do you have any reason to believe that it is even possible to rewrite the equation in such a way?
$endgroup$
– 5xum
Dec 12 '18 at 10:15
$begingroup$
Do you have any reason to believe that it is even possible to rewrite the equation in such a way?
$endgroup$
– 5xum
Dec 12 '18 at 10:15
$begingroup$
No I don't even know if you can, it's not a question from a textbook if that's what you're asking.
$endgroup$
– John Miller
Dec 12 '18 at 10:17
$begingroup$
No I don't even know if you can, it's not a question from a textbook if that's what you're asking.
$endgroup$
– John Miller
Dec 12 '18 at 10:17
$begingroup$
When one solves for $y$ using computer algebra, you get a horrendously complex solution.
$endgroup$
– David G. Stork
Dec 12 '18 at 10:17
$begingroup$
When one solves for $y$ using computer algebra, you get a horrendously complex solution.
$endgroup$
– David G. Stork
Dec 12 '18 at 10:17
$begingroup$
Oh why is this that some equations cannot be rearranged?
$endgroup$
– John Miller
Dec 12 '18 at 10:18
$begingroup$
Oh why is this that some equations cannot be rearranged?
$endgroup$
– John Miller
Dec 12 '18 at 10:18
$begingroup$
@JohnMiller That's, in a way, the wrong question to ask. The better question would be "why should all equations be such that they can be rearranged"?
$endgroup$
– 5xum
Dec 12 '18 at 10:19
$begingroup$
@JohnMiller That's, in a way, the wrong question to ask. The better question would be "why should all equations be such that they can be rearranged"?
$endgroup$
– 5xum
Dec 12 '18 at 10:19
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$x^4 - 2xyleft(xy + frac{k_1 k_2}{z}right) + y^4 = 0$
Rewriting a bit more:
$$color{blue}{y^4}-2x^2color{blue}{y^2}-2xfrac{k_1 k_2}{z}color{blue}{y}+x^4=0$$
This is a quartic equation in $y$ and in general, that's not easy to solve. Note that the term in $color{blue}{y^3}$ is missing and that makes it slightly better, you can take a look at Ferrari's solution (see depressed quartic equation for the equation without the third degree term) but it will get (very) ugly...
The question is, do you really need (want?) to symbolically solve this for $y$?
answered Dec 12 '18 at 10:22
StackTDStackTD
22.6k2050
$endgroup$
$begingroup$
The equation is for a magnet and I'm trying to find $y$ as that is the distance between, if they exist, the monopoles of a magnet.
$endgroup$
– John Miller
Dec 12 '18 at 10:24
$begingroup$
If you're not interested in an explicit formula for $y$ (which doesn't really give you a lot of insight anyway, if the formula is too ugly), you would typically use software te solve for $y$ with numerical methods.
$endgroup$
– StackTD
Dec 12 '18 at 10:25
$begingroup$
Yes I figured that would be easier, was just hoping I wouldn't have to ressort to that as it does not give as much insight as I wanted.
$endgroup$
– John Miller
Dec 12 '18 at 10:27
$begingroup$
You can always hope for a pretty solution :-). Now you know that sometimes that's not possible, sometimes it's ugly (too ugly to be useful?) and if you're lucky, you can get a nice explicit formula. I'd say you're in the middle case with this one.
$endgroup$
– StackTD
Dec 12 '18 at 10:29
$begingroup$
I've never heard of Ferrari's solution so I'll look into that for a bit, perhaps I'll get some inspiration but for a physical phenomenon it shouldn't be this 'ugly', at least it seems.
$endgroup$
– John Miller
Dec 12 '18 at 10:34
|
show 1 more comment
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$begingroup$
$x^4 - 2xyleft(xy + frac{k_1 k_2}{z}right) + y^4 = 0$
Rewriting a bit more:
$$color{blue}{y^4}-2x^2color{blue}{y^2}-2xfrac{k_1 k_2}{z}color{blue}{y}+x^4=0$$
This is a quartic equation in $y$ and in general, that's not easy to solve. Note that the term in $color{blue}{y^3}$ is missing and that makes it slightly better, you can take a look at Ferrari's solution (see depressed quartic equation for the equation without the third degree term) but it will get (very) ugly...
The question is, do you really need (want?) to symbolically solve this for $y$?
answered Dec 12 '18 at 10:22
StackTDStackTD
22.6k2050
$endgroup$
$begingroup$
The equation is for a magnet and I'm trying to find $y$ as that is the distance between, if they exist, the monopoles of a magnet.
$endgroup$
– John Miller
Dec 12 '18 at 10:24
$begingroup$
If you're not interested in an explicit formula for $y$ (which doesn't really give you a lot of insight anyway, if the formula is too ugly), you would typically use software te solve for $y$ with numerical methods.
$endgroup$
– StackTD
Dec 12 '18 at 10:25
$begingroup$
Yes I figured that would be easier, was just hoping I wouldn't have to ressort to that as it does not give as much insight as I wanted.
$endgroup$
– John Miller
Dec 12 '18 at 10:27
$begingroup$
You can always hope for a pretty solution :-). Now you know that sometimes that's not possible, sometimes it's ugly (too ugly to be useful?) and if you're lucky, you can get a nice explicit formula. I'd say you're in the middle case with this one.
$endgroup$
– StackTD
Dec 12 '18 at 10:29
$begingroup$
I've never heard of Ferrari's solution so I'll look into that for a bit, perhaps I'll get some inspiration but for a physical phenomenon it shouldn't be this 'ugly', at least it seems.
$endgroup$
– John Miller
Dec 12 '18 at 10:34
|
show 1 more comment
$begingroup$
$x^4 - 2xyleft(xy + frac{k_1 k_2}{z}right) + y^4 = 0$
Rewriting a bit more:
$$color{blue}{y^4}-2x^2color{blue}{y^2}-2xfrac{k_1 k_2}{z}color{blue}{y}+x^4=0$$
This is a quartic equation in $y$ and in general, that's not easy to solve. Note that the term in $color{blue}{y^3}$ is missing and that makes it slightly better, you can take a look at Ferrari's solution (see depressed quartic equation for the equation without the third degree term) but it will get (very) ugly...
The question is, do you really need (want?) to symbolically solve this for $y$?
answered Dec 12 '18 at 10:22
StackTDStackTD
22.6k2050
$endgroup$
$begingroup$
The equation is for a magnet and I'm trying to find $y$ as that is the distance between, if they exist, the monopoles of a magnet.
$endgroup$
– John Miller
Dec 12 '18 at 10:24
$begingroup$
If you're not interested in an explicit formula for $y$ (which doesn't really give you a lot of insight anyway, if the formula is too ugly), you would typically use software te solve for $y$ with numerical methods.
$endgroup$
– StackTD
Dec 12 '18 at 10:25
$begingroup$
Yes I figured that would be easier, was just hoping I wouldn't have to ressort to that as it does not give as much insight as I wanted.
$endgroup$
– John Miller
Dec 12 '18 at 10:27
$begingroup$
You can always hope for a pretty solution :-). Now you know that sometimes that's not possible, sometimes it's ugly (too ugly to be useful?) and if you're lucky, you can get a nice explicit formula. I'd say you're in the middle case with this one.
$endgroup$
– StackTD
Dec 12 '18 at 10:29
$begingroup$
I've never heard of Ferrari's solution so I'll look into that for a bit, perhaps I'll get some inspiration but for a physical phenomenon it shouldn't be this 'ugly', at least it seems.
$endgroup$
– John Miller
Dec 12 '18 at 10:34
|
show 1 more comment
$begingroup$
$x^4 - 2xyleft(xy + frac{k_1 k_2}{z}right) + y^4 = 0$
Rewriting a bit more:
$$color{blue}{y^4}-2x^2color{blue}{y^2}-2xfrac{k_1 k_2}{z}color{blue}{y}+x^4=0$$
This is a quartic equation in $y$ and in general, that's not easy to solve. Note that the term in $color{blue}{y^3}$ is missing and that makes it slightly better, you can take a look at Ferrari's solution (see depressed quartic equation for the equation without the third degree term) but it will get (very) ugly...
The question is, do you really need (want?) to symbolically solve this for $y$?
answered Dec 12 '18 at 10:22
StackTDStackTD
22.6k2050
$endgroup$
$x^4 - 2xyleft(xy + frac{k_1 k_2}{z}right) + y^4 = 0$
Rewriting a bit more:
$$color{blue}{y^4}-2x^2color{blue}{y^2}-2xfrac{k_1 k_2}{z}color{blue}{y}+x^4=0$$
This is a quartic equation in $y$ and in general, that's not easy to solve. Note that the term in $color{blue}{y^3}$ is missing and that makes it slightly better, you can take a look at Ferrari's solution (see depressed quartic equation for the equation without the third degree term) but it will get (very) ugly...
The question is, do you really need (want?) to symbolically solve this for $y$?
answered Dec 12 '18 at 10:22
StackTDStackTD
22.6k2050
answered Dec 12 '18 at 10:22
StackTDStackTD
22.6k2050
answered Dec 12 '18 at 10:22
StackTDStackTD
22.6k2050
answered Dec 12 '18 at 10:22
StackTDStackTD
22.6k2050
22.6k2050
$begingroup$
The equation is for a magnet and I'm trying to find $y$ as that is the distance between, if they exist, the monopoles of a magnet.
$endgroup$
– John Miller
Dec 12 '18 at 10:24
$begingroup$
If you're not interested in an explicit formula for $y$ (which doesn't really give you a lot of insight anyway, if the formula is too ugly), you would typically use software te solve for $y$ with numerical methods.
$endgroup$
– StackTD
Dec 12 '18 at 10:25
$begingroup$
Yes I figured that would be easier, was just hoping I wouldn't have to ressort to that as it does not give as much insight as I wanted.
$endgroup$
– John Miller
Dec 12 '18 at 10:27
$begingroup$
You can always hope for a pretty solution :-). Now you know that sometimes that's not possible, sometimes it's ugly (too ugly to be useful?) and if you're lucky, you can get a nice explicit formula. I'd say you're in the middle case with this one.
$endgroup$
– StackTD
Dec 12 '18 at 10:29
$begingroup$
I've never heard of Ferrari's solution so I'll look into that for a bit, perhaps I'll get some inspiration but for a physical phenomenon it shouldn't be this 'ugly', at least it seems.
$endgroup$
– John Miller
Dec 12 '18 at 10:34
|
show 1 more comment
$begingroup$
The equation is for a magnet and I'm trying to find $y$ as that is the distance between, if they exist, the monopoles of a magnet.
$endgroup$
– John Miller
Dec 12 '18 at 10:24
$begingroup$
If you're not interested in an explicit formula for $y$ (which doesn't really give you a lot of insight anyway, if the formula is too ugly), you would typically use software te solve for $y$ with numerical methods.
$endgroup$
– StackTD
Dec 12 '18 at 10:25
$begingroup$
Yes I figured that would be easier, was just hoping I wouldn't have to ressort to that as it does not give as much insight as I wanted.
$endgroup$
– John Miller
Dec 12 '18 at 10:27
$begingroup$
You can always hope for a pretty solution :-). Now you know that sometimes that's not possible, sometimes it's ugly (too ugly to be useful?) and if you're lucky, you can get a nice explicit formula. I'd say you're in the middle case with this one.
$endgroup$
– StackTD
Dec 12 '18 at 10:29
$begingroup$
I've never heard of Ferrari's solution so I'll look into that for a bit, perhaps I'll get some inspiration but for a physical phenomenon it shouldn't be this 'ugly', at least it seems.
$endgroup$
– John Miller
Dec 12 '18 at 10:34
$begingroup$
The equation is for a magnet and I'm trying to find $y$ as that is the distance between, if they exist, the monopoles of a magnet.
$endgroup$
– John Miller
Dec 12 '18 at 10:24
$begingroup$
The equation is for a magnet and I'm trying to find $y$ as that is the distance between, if they exist, the monopoles of a magnet.
$endgroup$
– John Miller
Dec 12 '18 at 10:24
$begingroup$
If you're not interested in an explicit formula for $y$ (which doesn't really give you a lot of insight anyway, if the formula is too ugly), you would typically use software te solve for $y$ with numerical methods.
$endgroup$
– StackTD
Dec 12 '18 at 10:25
$begingroup$
If you're not interested in an explicit formula for $y$ (which doesn't really give you a lot of insight anyway, if the formula is too ugly), you would typically use software te solve for $y$ with numerical methods.
$endgroup$
– StackTD
Dec 12 '18 at 10:25
$begingroup$
Yes I figured that would be easier, was just hoping I wouldn't have to ressort to that as it does not give as much insight as I wanted.
$endgroup$
– John Miller
Dec 12 '18 at 10:27
$begingroup$
Yes I figured that would be easier, was just hoping I wouldn't have to ressort to that as it does not give as much insight as I wanted.
$endgroup$
– John Miller
Dec 12 '18 at 10:27
$begingroup$
You can always hope for a pretty solution :-). Now you know that sometimes that's not possible, sometimes it's ugly (too ugly to be useful?) and if you're lucky, you can get a nice explicit formula. I'd say you're in the middle case with this one.
$endgroup$
– StackTD
Dec 12 '18 at 10:29
$begingroup$
You can always hope for a pretty solution :-). Now you know that sometimes that's not possible, sometimes it's ugly (too ugly to be useful?) and if you're lucky, you can get a nice explicit formula. I'd say you're in the middle case with this one.
$endgroup$
– StackTD
Dec 12 '18 at 10:29
$begingroup$
I've never heard of Ferrari's solution so I'll look into that for a bit, perhaps I'll get some inspiration but for a physical phenomenon it shouldn't be this 'ugly', at least it seems.
$endgroup$
– John Miller
Dec 12 '18 at 10:34
$begingroup$
I've never heard of Ferrari's solution so I'll look into that for a bit, perhaps I'll get some inspiration but for a physical phenomenon it shouldn't be this 'ugly', at least it seems.
$endgroup$
– John Miller
Dec 12 '18 at 10:34
|
show 1 more comment
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$begingroup$
Do you have any reason to believe that it is even possible to rewrite the equation in such a way?
$endgroup$
– 5xum
Dec 12 '18 at 10:15
$begingroup$
No I don't even know if you can, it's not a question from a textbook if that's what you're asking.
$endgroup$
– John Miller
Dec 12 '18 at 10:17
$begingroup$
When one solves for $y$ using computer algebra, you get a horrendously complex solution.
$endgroup$
– David G. Stork
Dec 12 '18 at 10:17
$begingroup$
Oh why is this that some equations cannot be rearranged?
$endgroup$
– John Miller
Dec 12 '18 at 10:18
$begingroup$
@JohnMiller That's, in a way, the wrong question to ask. The better question would be "why should all equations be such that they can be rearranged"?
$endgroup$
– 5xum
Dec 12 '18 at 10:19