Simple Nested Radical

$x=\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2..... } } } }$

$x=\sqrt { 2+x }$

${ x }^{ 2 }=2+x$

${ x }^{ 2 }-x-2=0$

$(x-2)(x+1)=0$

$x=2$ or $x=-1$

The value of $x$ cannot be negative (because of the definition of the square root function), so the solution is 2.

$x = \sqrt{ 2 +\sqrt{ 2 + \sqrt { 2 + \sqrt { 2 + ... } } } }$

$x^2 = 2 + \sqrt{ 2 +\sqrt{ 2 + \sqrt { 2 + \sqrt { 2 + ... } } } }$

Notice that:

$x^2 = 2 + x$

So:

$x^2 - x - 2 = 0$

Solving this equation you should get

$x = 2$ Or $x = -1$

However there is no square root whose result is less than 0 So, the only answer is:

$\boxed { x = 2}$

suppose, $x=\sqrt{2+\sqrt{2+\sqrt{2+.......}}}$

then, $x=\sqrt{2+x}$ ...................[infinitely going]

or, $x^2=2+x$

doing middle term,we get $x=2,-1$

but square root of positive value can not be negative.

so, $x=2$

$\begin{aligned} x & = & \sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2\dots } } } } \\ x & = & \sqrt { 2+x } \\ { x }^{ 2 } & = & 2+x \\ { x }^{ 2 }-x-2 & = & 0 \\ (x-2)(x+1) & = & 0 \\ x & = & \boxed { 2 } \quad (x>0) \end{aligned}$

For starters, the question stated, $x = \sqrt{2+\sqrt{2 +...}}$

While means the RHS (right hand side) is actually equal to $\sqrt{2 + x}$

$x = \sqrt{2 + x}$

$x² = 2 + x$

$x² - x - 2 = 0$

x = 2 or x = -1(NA)

So the answer is 2