# Find the value of $\sqrt{2+\sqrt{2+\sqrt{2+\dots}} }$ [duplicate]

How to prove that $\sqrt{2+\sqrt{2+\sqrt{2+\dots}} }=2$

## marked as duplicate by user147263, user99914, colormegone, Harish Chandra Rajpoot, KasterSep 18 '15 at 5:40

Let $\sqrt{2+\sqrt{2+...}}=a$. Then squaring both sides we have $a^2=2+a$. This is a quadratic equation in $a$. Find the solutions of $a$. Then $a=2$ or $a=-1$. Since $a>0, a=2$.
• The second solution to the equation $a^2=2+a$ is not $-2$ but $-1$. – Mercy King Sep 18 '15 at 5:30
• Hint: if $\sqrt{2} < x < 2$, show that $\sqrt{2} < \sqrt{2+x} < 2$ and that $2-\sqrt{2+x} < 2-x$. – marty cohen Sep 18 '15 at 18:35