Infinite nested radical with variables

Question

Let $a \bowtie b = a+\sqrt{b+\sqrt{b+\sqrt{b+...}}}$. If $4\bowtie y = 10$, find the value of $y$.

I know that $\sqrt{y+\sqrt{y+\sqrt{y+...}}}=6,$ but what do I do know? I know how to solve normal infinite nested radicals, but this is confusing me a bit...

Answer 1

Informal proof

Denote by $Y = \sqrt{y+\sqrt{y+\sqrt{y+...}}}$. Then you see that $Y^2-y=Y$. So if you know that $Y=6$, you deduce that $y =30$.

(To have a formal proof, one should have a precise definition - as a limit - of $ \sqrt{y+\sqrt{y+\sqrt{y+...}}}$)

Answer 2

Square the above equation. You will get $$y+\sqrt{y+\sqrt{y+...}}=36$$ or $y+6=36$. So $y=30$