Nested Radicals

The equation:

$\sqrt{6+\sqrt{6+\sqrt{6\cdots}}}=x$

Can be simplified to $\Rightarrow$ $\sqrt{6+x}=x$

So:

$(\sqrt{6+x})^2=x^2$ $\Rightarrow$ $6+x=x^2$ $\Rightarrow$ $x^2-x-6=0$

Then:

$(x-3)(x+2)$ Clearly $\Rightarrow$ $\boxed{x=3}$

Consider given =y then given can be written as (6+y)^(1/2)=y
Squaring both sides we have y^2-y-6=0 i.e
(y+2)(y-3)=0 i.e y=-2 and y=+3 rejecting -ve value we have y=given=3

Let , $\sqrt{6 + \sqrt{6 + \sqrt{6\cdots}}} = x \\ \Rightarrow \sqrt{6 + x} = x \\ \Rightarrow (\sqrt{6 + x})^2 = x^2 \\ \Rightarrow 6 + x = x^2 \\ \Rightarrow x^2 - x - 6 = 0 \\ \Rightarrow (x - 3)(x + 2) = 0 \\ \Rightarrow x = 3,-2 \\ So, x = \boxed3$

found this with this simple code .. double x=sqrt(6.0);

for(int i=100;i>=1;i--)
{
x= sqrt(6+x);
}

Let $x=\sqrt{6+\sqrt{6+\sqrt{6+\dotsm}}}$ Then $x=\sqrt{6+x}$ Simplifying this, we get: $x=\sqrt{6+x}$ $x^2=6+x$ $x^2-x-6=0$ Plugging the values in the quadratic formula, we get: $\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)}=\frac{1\pm\sqrt{1+24}}{2}=\frac{1\pm\sqrt{25}}{2}=\frac{1\pm5}{2}$ Solving this we get $x=-2\;or\;x=3$ as the question has asked for the positive value,so the answer is $\boxed{x=3}$