Square Root Tricks (2)

Let $\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5+\cdots}}}} = x$

$\implies \sqrt{5+x}=x$

Squaring both sides,

$x^2=5+x$

Re-arranging the terms,

$x^2-x-5=0$

Using the quadratic formula,

$x=\dfrac{1\pm\sqrt{21}}{2}$

Assuming $x$ to be positive,

$x=\dfrac{1+\sqrt{21}}{2}$

$x\approx2.79128784748\approx\boxed{2.79}$

Let the sum be $S$

Square it and you get that $S^2 = 5 + S$

So bt the quadratic formula, $S = ((1 + \sqrt{21})/2)$ = 2.79