What is the the square root of 6 + the square root of 6 + the square root of 6 +..............................

Question

so there is a square root over a 6+a square root which is over a 6+a square root........ it goes on to infinity

the answer is an interger

Answer 1

Suppose x=sqrt(6+[sqrt(6+......)])
and u=[ ]part i.e=sqrt(6+...........)
then
x=sqrt(6+u)
x^2=6+u
since this is an infinite series
we can suppose
u=x
then
x^2=6+x
x^2-x-6=0
x=3,-2

EDIT:
I want to add:
since the sign with sqrt is positive we take the positive sqrt
therefore the answer is 3

Answer 2

The square root of 6 can't be simplified, but the square root of 12 can. √12 = √(4 × 3) = √4 × √3 = 2√3

Answer 3

if you want the square root of six, it is something around 2.44 and if you continue to add the square root of six plus root six plus root six to infinity, the answer is infinity. i think you perhaps have not written the problem out correctly. rewrite it and if i'm still on i'll help you out then.

Answer 4

The former answer is right except that -2 is of course NOT a solution. So your answer is 3.

Answer 5

The answer is not one but two integers (-3 and +2). It can be solved like a quadratic equation.

Answer 6

the former answer: that x = -2 is NOT a solution is invalid

- 2 also qualifies

Answer 7

I doubt it