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Hello I'm wondering whether anyone can help me rearrange this to solve for $x$, where $a$, $b$, $c$ are constants. I initially thought about a trig substitution $x=b\cdot \cos(z)$ but that just lead to another brick wall. Any insights gratefully appreciated!

Thanks Guy

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Put $a^2-x^2=z^2$, you will get the same 4 degree polynomial but it will not be as big as this. I am trying to solve it.

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Good one. No, I don't think a typical pre-calculus student can solve the resulting quartic. – Harald Hanche-Olsen Aug 15 '12 at 12:45
It is:z^4-2cz^3+b^2z-a^2z-b^c+a^c=0 – SN77 Aug 15 '12 at 12:50
@SN77: Surely, you did not intend to take the $c$th power of anything? I end up with $z^4-2cz^3+(b^2-a^2)z^2-2c(b-a)z+(b^2-a^2)c=0$ myself, but I don't have the time to do it really carefully. – Harald Hanche-Olsen Aug 15 '12 at 13:11
@picakhu: Doesn't look quadratic to me. – Harald Hanche-Olsen Aug 15 '12 at 13:16
Thanks for the quick help @harald-hanche-olsen. z^4-2cz^3+(b^2-a^2)z^2-2c(b^2-a^2)z+(b^2-a^2)c=0 – Guy Aug 15 '12 at 13:47

Put $1/c$ on one side of the equation, square both sides, separate out the square root and do again so that

$x$ is a root of $$\begin{multline}x^8+(-2b^2-2a^2+2c^2+2a^2c^2)x^6\\+(-4b^2a^2c^2+a^4-2a^4c^2-4a^2c^2-2b^2c^2+a^4c^4\\+b^4+4b^2a^2-2a^2c^4+c^4)x^4\\+(4b^2a^2c^2+2b^2a^2c^4-2a^4c^4b^2+4a^4c^2b^2+2b^4a^2c^2-2b^4a^2\\-2a^4b^2-2a^2c^4+2a^4c^2+2a^4c^4)x^2\\-2a^4c^4b^2+b^4a^4c^4-2b^4a^4c^2-2a^4c^2b^2+b^4a^4+a^4c^4\end{multline}$$

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solving a 4th order polynomial isn't going to be easy... – picakhu Aug 15 '12 at 12:35
Oh, didn't see the $x^8$ - line was too long. New edit makes it clearer. – Thomas Andrews Aug 15 '12 at 16:22

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