# Square root till infinity

What is the value of $\sqrt{x + \sqrt{ x + \sqrt{ x + \cdots } } }\,$? I know the basic trick to calculate this using $f = \sqrt{ x + f }$. But, I want more accurate answer which is I am not getting with this formula.

Squaring we get $f^2=x+f\iff f^2-f-x=0\implies f=\dfrac{1\pm\sqrt{1+4x}}2$
Now as $f>0,$ discard the negative root assuming $x>0$
• @JackD'Aurizio, Thanks. How to discard $$\frac{1-\sqrt{1+4x}}2$$ if $x<0$? – lab bhattacharjee Sep 20 '14 at 4:50
If $f=\sqrt{x+f}$ then $f^2-f-x=0$, hence $$f=\frac{1+\sqrt{1+4x}}{2}.$$