If a and b are positive real numbers, then
- $\sqrt[2]{\frac{a+\sqrt[2]{a^2-b}}{2}}+\sqrt[2]{\frac{a-\sqrt[2]{a^2-b}}{2}} = \sqrt[2]{a+\sqrt[2]{b}}$
Any idea about how proof this formula is welcome.
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egarro
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See this answer. – dxiv Mar 03 '17 at 02:49
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I found this: https://it.wikipedia.org/wiki/Radicale_doppio, but I'm not get it how the solution of a symmetrical equation system can be found with a quadratic equation. – egarro Mar 03 '17 at 03:31
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Hint: let $\;u=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\;$ and $\;v=\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\,$. Then:
$\require{cancel} u^2+v^2=\frac{a+\bcancel{\sqrt{a^2-b}}}{2}+\frac{a-\bcancel{\sqrt{a^2-b}}}{2}=\frac{\cancel{2}a}{\cancel{2}}=a$
$u\,v = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \sqrt{\frac{a-\sqrt{a^2-b}}{2}}=\sqrt{\frac{\cancel{a^2}-(\cancel{a^2}-b)}{4}}=\frac{\sqrt{b}}{2}$
Therefore $\;(u+v)^2=u^2+v^2+2uv=a+\sqrt{b}\;$ which proves the posted identity.
Similarly, $\;(u-v)^2=u^2+v^2-2uv=a-\sqrt{b}\;$ which proves the related identity:
$$\sqrt{\frac{a+\sqrt{a^2-b}}{2}} - \sqrt{\frac{a-\sqrt{a^2-b}}{2}} = \sqrt{a-\sqrt{b}}$$
dxiv
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