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PostPosted: Thu, 27 Oct 2011 22:50:02 UTC 
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My teacher today put the following problem on the board:

x^2+2x-2y^3+y=5

She asked us to solve for y and said it could not be done. I'm fairly confident that it does in fact have an answer. If somebody could explain how this problem could be solved and work it out with steps to prove to her I would be grateful. :)

Thanks to any who contributes. :D


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PostPosted: Fri, 28 Oct 2011 00:00:03 UTC 
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Obviously you can't get a specific number for y. However, it is a cubic polynomial in y, which is solvable (although the general solution is not easy). The net result will be 3 y's as functions of x.


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PostPosted: Fri, 28 Oct 2011 00:41:45 UTC 
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I appreciate the reply. How would I go about solving for the exact value of y in this case?


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PostPosted: Fri, 28 Oct 2011 01:24:34 UTC 
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Ckrupala wrote:
My teacher today put the following problem on the board:

x^2+2x-2y^3+y=5

She asked us to solve for y and said it could not be done. I'm fairly confident that it does in fact have an answer. If somebody could explain how this problem could be solved and work it out with steps to prove to her I would be grateful. :)

Thanks to any who contributes. :D


Of course it cannot be done, that's the equation for a curve in the plane, there are multiple points that work since it's a cubic in y, there is no unique solution, there is no way to make this a pure function in just y and still get the same plane curve. There is provable no way to succeed, so you're out of luck.

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PostPosted: Fri, 28 Oct 2011 03:35:37 UTC 
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This is the supposed answer: y = -(-108 x^2+sqrt((-108 x^2-216 x+540)^2-864)-216 x+540)^(1/3)/(6 2^(1/3))-2^(1/3)/(-108 x^2+sqrt((-108 x^2-216 x+540)^2-864)-216 x+540)^(1/3)

Any thoughts on the answer above?


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PostPosted: Fri, 28 Oct 2011 04:49:10 UTC 
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Ckrupala wrote:
This is the supposed answer: y = -(-108 x^2+sqrt((-108 x^2-216 x+540)^2-864)-216 x+540)^(1/3)/(6 2^(1/3))-2^(1/3)/(-108 x^2+sqrt((-108 x^2-216 x+540)^2-864)-216 x+540)^(1/3)

Any thoughts on the answer above?


This is just one of the three roots out there. Have a look at S.O.S. Math's The Geometry of the Cubic Formula.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Fri, 28 Oct 2011 04:52:28 UTC 
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Go enter your equation here:
http://www.numberempire.com/equationsolver.php

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PostPosted: Fri, 28 Oct 2011 05:13:36 UTC 
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Denis wrote:


That's a fancy doo-hickey you have there Denis.

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PostPosted: Fri, 28 Oct 2011 06:04:29 UTC 
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So technically the 3 solutions I get from entering that equation would be the answer.

That link is quite snazzy.


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PostPosted: Fri, 28 Oct 2011 06:47:17 UTC 
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Ckrupala wrote:
So technically the 3 solutions I get from entering that equation would be the answer.

That link is quite snazzy.


No, solving for y means you can pin down what y is given what x is, you have three possibilities, which means there is no way to determine y given x, only a set of possibilities.

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PostPosted: Fri, 28 Oct 2011 19:10:45 UTC 
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Take something REALLY simple, like y = 2x + 1
Solutions are limitless (there is one for ANY value assigned to x), right?
Your problem is no different...

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PostPosted: Fri, 28 Oct 2011 19:15:09 UTC 
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Shadow wrote:
That's a fancy doo-hickey you have there Denis.

Ya; try a*x^3 + b*x^2 + c*x + d = 0

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PostPosted: Fri, 28 Oct 2011 19:15:37 UTC 
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Denis wrote:
Shadow wrote:
That's a fancy doo-hickey you have there Denis.

Ya; try a*x^3 + b*x^2 + c*x + d = 0


:)

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