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Solving $KL^2C - LC^2C = (B-C)C^2C - C^2C(T+KG^2) - KP^2$ for $L$
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How can I solve this equation for $L$. Everything else is known number: $$KL^2C - LC^2C = (B-C)C^2C - C^2C(T+KG^2) - KP^2$$
exponentiation
edited Jul 30 at 20:35
Blue
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asked Jul 30 at 20:28
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up vote
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down vote
favorite
How can I solve this equation for $L$. Everything else is known number: $$KL^2C - LC^2C = (B-C)C^2C - C^2C(T+KG^2) - KP^2$$
exponentiation
edited Jul 30 at 20:35
Blue
43.6k868141
asked Jul 30 at 20:28
ÃÂøúþûðù Úðрðôöþò
1
1
Well, what is $C$? if $C=1$ this is just a quadratic equation in $L$. If $C$ is a large integer then this is a hard polynomial to solve. But maybe $C$ isn't even a positive integer?
– lulu
Jul 30 at 20:32
1
I think you are out of luck unless $2C$ is $1$ or $2$ or $3$ or $4$. Numerical methods may work when that's not the case.
– Ethan Bolker
Jul 30 at 20:32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How can I solve this equation for $L$. Everything else is known number: $$KL^2C - LC^2C = (B-C)C^2C - C^2C(T+KG^2) - KP^2$$
exponentiation
edited Jul 30 at 20:35
Blue
43.6k868141
asked Jul 30 at 20:28
ÃÂøúþûðù Úðрðôöþò
1
How can I solve this equation for $L$. Everything else is known number: $$KL^2C - LC^2C = (B-C)C^2C - C^2C(T+KG^2) - KP^2$$
exponentiation
edited Jul 30 at 20:35
Blue
43.6k868141
asked Jul 30 at 20:28
ÃÂøúþûðù Úðрðôöþò
1
edited Jul 30 at 20:35
Blue
43.6k868141
edited Jul 30 at 20:35
Blue
43.6k868141
edited Jul 30 at 20:35
Blue
43.6k868141
43.6k868141
asked Jul 30 at 20:28
ÃÂøúþûðù Úðрðôöþò
1
asked Jul 30 at 20:28
ÃÂøúþûðù Úðрðôöþò
1
asked Jul 30 at 20:28
ÃÂøúþûðù Úðрðôöþò
1
1
1
Well, what is $C$? if $C=1$ this is just a quadratic equation in $L$. If $C$ is a large integer then this is a hard polynomial to solve. But maybe $C$ isn't even a positive integer?
– lulu
Jul 30 at 20:32
1
I think you are out of luck unless $2C$ is $1$ or $2$ or $3$ or $4$. Numerical methods may work when that's not the case.
– Ethan Bolker
Jul 30 at 20:32
add a comment |Â
1
Well, what is $C$? if $C=1$ this is just a quadratic equation in $L$. If $C$ is a large integer then this is a hard polynomial to solve. But maybe $C$ isn't even a positive integer?
– lulu
Jul 30 at 20:32
1
I think you are out of luck unless $2C$ is $1$ or $2$ or $3$ or $4$. Numerical methods may work when that's not the case.
– Ethan Bolker
Jul 30 at 20:32
1
1
Well, what is $C$? if $C=1$ this is just a quadratic equation in $L$. If $C$ is a large integer then this is a hard polynomial to solve. But maybe $C$ isn't even a positive integer?
– lulu
Jul 30 at 20:32
Well, what is $C$? if $C=1$ this is just a quadratic equation in $L$. If $C$ is a large integer then this is a hard polynomial to solve. But maybe $C$ isn't even a positive integer?
– lulu
Jul 30 at 20:32
1
1
I think you are out of luck unless $2C$ is $1$ or $2$ or $3$ or $4$. Numerical methods may work when that's not the case.
– Ethan Bolker
Jul 30 at 20:32
I think you are out of luck unless $2C$ is $1$ or $2$ or $3$ or $4$. Numerical methods may work when that's not the case.
– Ethan Bolker
Jul 30 at 20:32
add a comment |Â
1 Answer
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If $K = 0$ this is a line equation in $L$, so let's assume $K ne 0$. Let's also assume $C$ is a non-negative integer.
Then you are looking at $$L^2C - aL -b = 0$$ and for $C>2$ you will not necessarily have analytic solutions.
For $C=0$, it is also linear in $L$. For $C=1$, the quadratic formula applies, and for $C=2$, you have to solve a 4-degree equation, for which you can find closed analytic formulae akin to the quadratic formula...
answered Jul 30 at 20:34
gt6989b
30.2k22148
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If $K = 0$ this is a line equation in $L$, so let's assume $K ne 0$. Let's also assume $C$ is a non-negative integer.
Then you are looking at $$L^2C - aL -b = 0$$ and for $C>2$ you will not necessarily have analytic solutions.
For $C=0$, it is also linear in $L$. For $C=1$, the quadratic formula applies, and for $C=2$, you have to solve a 4-degree equation, for which you can find closed analytic formulae akin to the quadratic formula...
answered Jul 30 at 20:34
gt6989b
30.2k22148
add a comment |Â
up vote
1
down vote
If $K = 0$ this is a line equation in $L$, so let's assume $K ne 0$. Let's also assume $C$ is a non-negative integer.
Then you are looking at $$L^2C - aL -b = 0$$ and for $C>2$ you will not necessarily have analytic solutions.
For $C=0$, it is also linear in $L$. For $C=1$, the quadratic formula applies, and for $C=2$, you have to solve a 4-degree equation, for which you can find closed analytic formulae akin to the quadratic formula...
answered Jul 30 at 20:34
gt6989b
30.2k22148
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $K = 0$ this is a line equation in $L$, so let's assume $K ne 0$. Let's also assume $C$ is a non-negative integer.
Then you are looking at $$L^2C - aL -b = 0$$ and for $C>2$ you will not necessarily have analytic solutions.
For $C=0$, it is also linear in $L$. For $C=1$, the quadratic formula applies, and for $C=2$, you have to solve a 4-degree equation, for which you can find closed analytic formulae akin to the quadratic formula...
answered Jul 30 at 20:34
gt6989b
30.2k22148
If $K = 0$ this is a line equation in $L$, so let's assume $K ne 0$. Let's also assume $C$ is a non-negative integer.
Then you are looking at $$L^2C - aL -b = 0$$ and for $C>2$ you will not necessarily have analytic solutions.
For $C=0$, it is also linear in $L$. For $C=1$, the quadratic formula applies, and for $C=2$, you have to solve a 4-degree equation, for which you can find closed analytic formulae akin to the quadratic formula...
answered Jul 30 at 20:34
gt6989b
30.2k22148
answered Jul 30 at 20:34
gt6989b
30.2k22148
answered Jul 30 at 20:34
gt6989b
30.2k22148
answered Jul 30 at 20:34
gt6989b
30.2k22148
30.2k22148
add a comment |Â
add a comment |Â
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1
Well, what is $C$? if $C=1$ this is just a quadratic equation in $L$. If $C$ is a large integer then this is a hard polynomial to solve. But maybe $C$ isn't even a positive integer?
– lulu
Jul 30 at 20:32
1
I think you are out of luck unless $2C$ is $1$ or $2$ or $3$ or $4$. Numerical methods may work when that's not the case.
– Ethan Bolker
Jul 30 at 20:32