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I want to know how to solve nested radicals of the following form:

$$ \sqrt{a-y\sqrt{a+y\sqrt{a-y\sqrt{\cdots}}}} $$

The repetition of signs is $ -,+,- $

Mrigank Shekhar Pathak
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    This - https://math.stackexchange.com/questions/1596824/solution-of-nested-radical-sqrt72-sqrt7-2-sqrt7-2x-x?rq=1 - might help. – rookie Jun 12 '17 at 07:14

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Set $$x = \sqrt{a-y\sqrt{a+y\sqrt{a-y\sqrt{\cdots}}}}.$$ Then you have $$x = \sqrt{a-y\sqrt{a+yx}}.$$ Squaring both sides and collecting terms: $$x^2 - a = -y\sqrt{a+yx}.$$ Squaring both sides again: $$(x^2 - a)^2 = {y^2}(a + yx).$$ It remains to solve for $x$. Can you take it from here?

Jose Arnaldo Bebita Dris
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In addition to JOSE ARNALDO BEBITA DRIS post, we can solve the equation pretending that the $a$ is variable and solving quadratic equation for it. Then we will find $x$ in terms of $a$ and $y$. The result is: $$x = \frac{-y\pm\sqrt{4 a - 3 y^2}}{2} \text{ or } x=\frac{y\pm\sqrt{4 a + y^2} }{2}.$$ Plug each obtained solution into the original equation and one of them will be the exact solution to our nested root problem or just notice that only root can be $x=\dfrac{-y+\sqrt{4 a - 3y^2}}{2}$.

Mr_Fox_AkA_Ant
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