Dear Tejas, challenge ACCOMPLISHED! (Speed run, Uncut)

What people said

Recommends For the love of maths · 2018-10 · Mathematics Stack Exchange · on “How to prove $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$”

So, I was watching this [this video] video by blackpenredpen [LINK] where he mentions that $$\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=\frac {\sqrt {4a-3}-1}2$$ so I wanted to try and prove it myself. Let $\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}}=x$ But $\sqrt {a-\sqrt {a+\underbrace{\sqrt {a-\sqrt {a+\sqrt {a-\sqrt {a+\ldots}}}}}_x}}=x$ $\therefore \sqrt {a-\sqrt {a+x}}=x$ $a-\sqrt {a+x}=x^2$ $x^2-a=-\sqrt {a+x}$ $x^4-2ax^2+a^2=a+x$ $x^4-2ax^2-x+a^2-a=0$ Note that this is of the form $y^4+py^2+qy+r=0$ so we can use Ferrari-Cardano [LINK] . We need to find a $z$ such that $(2z-p)y^2-qy+(z^2-r)$ has a discriminant of $0$ . The discriminant of $(2z

Recommends Giraffes4thewin · 2020-04 · Mathematics Stack Exchange · on “A possible solution to $\sqrt {5-x}=5-x^2$ (without taking square from both sides)”

I can’t answer your question, but redpenblackpen posted a video on this exact problem you might want to check out. Here’s the link - [this video]

Recommends Aiden Chow · 2020-06 · Mathematics Stack Exchange · on “How to make inverse to work here?”

I found a [this video] video by a YouTuber named blackpenredpen [LINK] that explains how to solve for the second solution that you are seeking. He basically uses infinitely nested square roots instead of inverses but you might find it helpful.