# Question 47095

Mar 3, 2016

Sorry for the inconvenience
(this answer is based on a mathematical trick and not explanation )

#### Explanation:

$A n s w e r = 2$

The trick for this:

If you see a question lie adding the roots of roots of a same number to infinite times,Take out the number in it (in this case is $2$)

And make the number in the form

$n \left(n + 1\right) = 2$

If we solve for it, we get $n = 1 , n + 1 = 2$

So,$n + 1$ will be the solution

Some examples

$\sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12}}}} \ldots = 4$

$\sqrt{72 + \sqrt{72 + \sqrt{72 + \sqrt{72}}}} \ldots = 9$

Mar 3, 2016

sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2 + … ∞)))) = 2

#### Explanation:

sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2 + … ))))

This is an amazing concept which is quite simple and easy
So look carefully
Lets say
sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2 + … ∞))) )= x

Now square both sides

2 +color(red)(" sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2 + … ∞))) )) = x^2

Now if you notice the part under our radical is the same as under our first equation;

color(red)sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2 + … ∞)))) = x

Lets plug the x where it belong in the second equation

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

Right now i think we are getting somewhere

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

Factor this

$\left(x - 2\right) \left(x + 1\right) = 0$

$x = 2 \mathmr{and} - 1$

Lets remind ourselves

sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2 + … ∞)))) = x

Clearly this is positive

So $x \ne - 1$

Hence

sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2 + … ∞)))) = x = 2#