Question #47095

2 Answers
Mar 3, 2016

Answer:

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

Explanation:

#Answer=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(n+1)=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))))...=4#

#sqrt(72+sqrt(72+sqrt(72+sqrt(72))))...=9#

Mar 3, 2016

Answer:

#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

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

#x = 2 or -1#

Lets remind ourselves

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

Clearly this is positive

So #x != -1#

Hence

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