Question

What is `sqrt(7 + sqrt(7 - sqrt (7 + sqrt (7 - sqrt(7 + ...... ∞`?

Answer 1

`3`

Explanation:

Let

`x=sqrt(7+sqrt(7-sqrt(7+sqrt(7-sqrt(7-sqrt(7+...oo`

where we constrain our solution to be positive since we are taking only the positive square root i.e. `x>=0`. Squaring both sides we have

`x^2=7+sqrt(7-sqrt(7+sqrt(7-sqrt(7-sqrt(7+...oo`

`=>x^2-7=sqrt(7-sqrt(7+sqrt(7-sqrt(7-sqrt(7+...oo`

Where this time we constrain the left hand side to be positive, since we only want the positive square root i.e.

`x^2-7>=0` `=>` `x>=sqrt(7) ~= 2.65`

where we have eliminated the possibility the `x<=-sqrt(7)` using our first constraint.

Again squaring both sides we have

`(x^2-7)^2`=`7-sqrt(7+sqrt(7-sqrt(7-sqrt(7+........oo`

`(x^2-7)^2-7=-sqrt(7+sqrt(7-sqrt(7-sqrt(7+........oo`

The expression in the repeated square roots is the original expression for `x`, therefore

`(x^2-7)^2-7=-x`

or

`(x^2-7)^2-7+x=0`

Trial solutions of this equation are `x=-2` and `x=+3` which results in the following factorization

`(x+2)(x-3)(x^2+x-7)=0`

Using the quadratic formula on the third factor `(x^2+x-7)=0` gives us two more roots:

`(-1+-sqrt(29))/2 ~= 2.19 " and " -3.19`

The four roots of the polynomial are therefore `-3.19..., -2, 2.19...,` and `3`. Only one of these values satisfies our constraint `x>=sqrt(7) ~= 2.65`, therefore

`x=3`

Answer 2

Another way

Explanation:

I like to discuss a tricky way to have a solution at a glance on the problem of repeated square roots like the following
`sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo`

where `r` belongs to the following series
`3,7,13,21,31............`, the general term of which is given by
`m^2-m+1` where `m epsilon N` and `m>1`

TRICK
If 1 is subtracted from the given Number `m^2-m+1` the resulting number becomes `m^2-m` which is `m(m-1)` and which is nothing but the product of two consecutive number and larger one of these two will be the unique solution of the problem.

when r = `m^2-m+1` the factor of `m^2-m+1-1` = `(m-1)m` and m is the answer

when r = 3 the factor of (3-1)= 2 = 1.2 and 2 is the answer
when r = 7 the factor of (7-1) =6= 2.3 and 3 is the answer
and so on.......

Explanation
Taking
`x=sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo`
Squaring both sides
`x^2= r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo`

`x^2- r=sqrt(r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo`
Again Squaring both sides
`(x^2- r)^2=r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo`
`(x^2- r)^2-r=-x`
`(x^2- r)^2-r+x=0`
putting r = `m^2-m+1`

`(x^2- (m^2-m+1))^2-(m^2-m+1)+x=0`

if we put x = m in the LHS of this equation the LHS becomes

LHS =
`(m^2- (m^2-m+1))^2-(m^2-m+1)+m`
`=(cancel(m^2)- cancel(m^2)+m-1))^2-(m^2-m+1-m)`

`=(m-1))^2-(m-1)^2=0`
the equation is satisfied.
Hence m is the answer

Answer 3

let's put

`x=sqrt(7+sqrt(7- sqrt(7+sqrt(7-sqrt....`

We can easily see that

`sqrt(7+sqrt(7-x))=x`

So let's solve the equation:

`7+sqrt(7-x)=x^2`

`sqrt(7-x)=x^2-7`

`7-x=(x^2-7)^2=x^4-14x^2+49`

`x^4-14x^2+x +42=0`

This is not a trivial equation to be solved. One of the other persons that answered the question referred the solution 3. If you try it, you will see it's true.