<s>[INST] What $x$ solves $x^2 = R^2 - x^2 +x(R^2 - x^2)^\frac{1}{2}$? This is from "Calculus Made Easy", Exercise 9 problem 9c.  The formula in the title is what I've gotten and I know from the answer in the back of the book that it reduces to $x = 0.8507R$, but I can't see how to reduce it further (other than grouping the two $x^2$ terms).
By the way, the formula results from differentiating $4\pi x(R^2-x^2)^\frac{1}{2}+2\pi x^2$ and setting to zero.
 [/INST] Rearrange to get $$\frac{2x^2-R^2}{x} = (R^2-x^2)^{\frac{1}{2}}$$
Square both sides: $$\frac{4x^4-4x^2R^2+R^4}{x^2} = R^2-x^2$$
Multiply by $x^2$: $$4x^4-4x^2R^2+R^4 = x^2R^2-x^4$$
Rearrange some more: $$5x^4-5x^2R^2+R^4 = 0$$
This is a quadratic in $x^2$: $$x^2 = \frac{5R^2\pm \sqrt{25R^4-20R^4}}{10} = \frac{5R^2\pm R^2\sqrt{5}}{10} = (\frac{5 \pm \sqrt{5}}{10})R^2$$
Taking the square root finds $$x = \pm R\sqrt{\frac{5 \pm \sqrt{5}}{10}}$$
The non-extraneous solutions are $$\pm R\sqrt{\frac{5+\sqrt{5}}{10}} \approx \pm 0.8507R$$ for $x \ge 0$.
For $x \le 0$ the non-extraneous solutions are $$\pm R\sqrt{\frac{5-\sqrt{5}}{10}} \approx \pm 0.5257R$$
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