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Consider the following equations: $$1-\frac{\mu}{1+x_{1}-x_{2}^{2}} = 0 \\ -2 + \frac{2 \mu x_{2}}{1+x_{1}-x_{2}^{2}}-\frac{\mu}{x_2} = 0$$

Assume that $1+x_1-x_{2}^{2} \geq 0$ and $x_{2} \geq 0$. Solving for $x_2$ in terms of $\mu$ is just a matter of back substitution? In other words using the first equation to solve for $x_1$ in terms of $x_2$. Then substitute this into the second equation to get a quadratic in $x_2$?

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You use the first equation to simplify the second, leading to a quadratic in $x_2$ and $\mu$ (no $x_1$s anymore); then the inequalities let you figure out which of the two values is $x_2$. See below. –  Arturo Magidin May 9 '12 at 2:04

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From the first equation we obtain $$\begin{align*} 1 - \frac{\mu}{1+x_1-x_2^2} &= 0\\ 1 &= \frac{\mu}{1+x_1-x_2^2} \end{align*}$$

You can then simplify the second equation by noting that the second summand gives $$\frac{2\mu x_2}{1+x_1-x_2^2} = \frac{\mu}{1+x_1-x_2^2}\left(2x_2\right) = 2x_2,$$ which means that the second equation is equivalent to $$-2 + 2x_2 - \frac{\mu}{x_2} = 0$$ and we can solve this for $x_2$ in terms of $\mu$ by multiplying through by $x_2$ and solving the quadratic: $$\begin{align*} -2 + 2x_2 - \frac{\mu}{x_2} &= 0\\ -2x_2 + 2x_2^2 - \mu &= 0\\ x_2 &= \frac{2\pm\sqrt{4 + 8\mu}}{4}\\ &= \frac{2\pm 2\sqrt{1+2\mu}}{4}\\ &= \frac{1\pm\sqrt{1+2\mu}}{2}. \end{align*}$$ Since $\mu\gt 0$, $\sqrt{1+2\mu}\gt 1$, hence $1-\sqrt{1+2\mu}\lt 0$. So we conclude that $x_2 = \frac{1}{2}(1 + \sqrt{1+2\mu})$.

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