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

Mar 9, 2016

$3$

#### 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. $x \ge 0$. 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.

${x}^{2} - 7 \ge 0$ $\implies$ $x \ge \sqrt{7} \cong 2.65$

where we have eliminated the possibility the $x \le - \sqrt{7}$ using our first constraint.

Again squaring both sides we have

${\left({x}^{2} - 7\right)}^{2}$=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

${\left({x}^{2} - 7\right)}^{2} - 7 = - x$

or

${\left({x}^{2} - 7\right)}^{2} - 7 + x = 0$

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

$\left(x + 2\right) \left(x - 3\right) \left({x}^{2} + x - 7\right) = 0$

Using the quadratic formula on the third factor $\left({x}^{2} + x - 7\right) = 0$ gives us two more roots:

$\frac{- 1 \pm \sqrt{29}}{2} \cong 2.19 \text{ and } - 3.19$

The four roots of the polynomial are therefore $- 3.19 \ldots , - 2 , 2.19 \ldots ,$ and $3$. Only one of these values satisfies our constraint $x \ge \sqrt{7} \cong 2.65$, therefore

$x = 3$

Mar 10, 2016

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
$3 , 7 , 13 , 21 , 31. \ldots \ldots \ldots . .$, 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 $m \left(m - 1\right)$ 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 ${m}^{2} - m + 1 - 1$ = $\left(m - 1\right) m$ 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
${\left({x}^{2} - r\right)}^{2} - r = - x$
${\left({x}^{2} - r\right)}^{2} - r + x = 0$
putting r = ${m}^{2} - m + 1$

${\left({x}^{2} - \left({m}^{2} - m + 1\right)\right)}^{2} - \left({m}^{2} - m + 1\right) + x = 0$

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

LHS =
${\left({m}^{2} - \left({m}^{2} - m + 1\right)\right)}^{2} - \left({m}^{2} - m + 1\right) + m$
=(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.

Mar 10, 2016

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:

$7 + \sqrt{7 - x} = {x}^{2}$

$\sqrt{7 - x} = {x}^{2} - 7$

$7 - x = {\left({x}^{2} - 7\right)}^{2} = {x}^{4} - 14 {x}^{2} + 49$

${x}^{4} - 14 {x}^{2} + x + 42 = 0$

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.