Rooting All The Way

$Let\quad x=\sqrt { 6+\sqrt { 6+\sqrt { 6+... } } } \\ { \quad \quad x }^{ 2 }=6+\sqrt { 6+\sqrt { 6+\sqrt { 6+... } } } \\ Since\quad x=\sqrt { 6+\sqrt { 6+\sqrt { 6+... } } } \\ \quad \quad { x }^{ 2 }=6+x\\ \quad { x }^{ 2 }-x-6=0\\ (x-3)(x+2)=0\\ \quad x=3\quad (x>0)\\ \\ Therefore,\quad \\ \boxed { \sqrt { 6+\sqrt { 6+\sqrt { 6+... } } } =3 }$

Let $x = \sqrt {6 + \sqrt {6 + \sqrt {6 + \ldots}}}$

Since $\sqrt {n} = n^\frac {1}{2}$

$\sqrt {n}^2 = n$

$x^2 = \sqrt {6 + \sqrt {6 + \sqrt {6 + \ldots}}}^2 = 6 + \sqrt {6 + \sqrt {6 + \sqrt {6 + \ldots}}}$

If we then take $x$ from $x^2$ we get

$x^2 - x = 6$

This can be re-written as a quadratic equation

$x^2 - x - 6 = 0$

Putting this into the quadratic formula we get

$x = \frac {1 \pm \sqrt {(-1)^2 - 4(1)(-6)}}{2}$

This simplifies down to

$x = \frac {1 \pm \sqrt {25}}{2}$

Which then means that

$x = \frac {1 \pm 5}{2}$

This gives a positive and negative value of $x$

$x = 3$ and $x = -2$

Since the question asked for a positive value the answer is $3$

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