$\sqrt { x\pm \sqrt { x\pm \sqrt { x\pm \sqrt { x\pm \cdots } } } }$

Within this infinitely nested radical, exactly 2018 of the $\pm$ symbols are assigned to be minus $(-)$ signs, with the rest of the $\pm$ symbols assigned to be plus $(+)$ signs (the minus signs don't need to be consecutive).

What's the
**
minimum positive value
**
of
$x$
that guarantees the above expression has a real value?

Enter your answer in the form $\left\lfloor 1000x \right\rfloor.$

**
Note:
**
Consider the worst-case scenario, among all possible places where the
$2018$
minuses could be.

The answer is 2000.

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If there is a finite number of minuses in an infintite sequence of $+$ 's and $-$ 's, then after the last minus there is an embedded infinite nested radical consisting only of positive $x$ 's The value of this can be found by solving for $z$

$z = \sqrt{x+z}$

the resulting being $z = \dfrac{1}{2}\left(1+\sqrt{1+4x}\right)$

For the radical $\sqrt{x - z}$ to be real, $x \ge z$ , so that solving for $x$ here

$x = \dfrac{1}{2}\left(1+\sqrt{1+4x}\right)$

gets us $x \ge 2$ . Otherwise if $x<2$ , this radical becomes imaginary.

Suppose with the last minus in the sequence, the radical becomes imaginary. Thereafter, in going back outwards towards less embedded radicals, whenever a complex number is added or substracted from a real number ( $x$ in this case), the radical of the sum or difference is still imaginary. Thus, the entire infinite nested radical is imaginary, once any of its embedded radicals becomes imaginary. It may be possible that as a limit, some infinite nested radical beginning with an imaginary number in its most embedded radical "may approach a real number", but there is only a finite number of radicals from the last minus to the start.

If $x \ge 2$ , then the radical containing the last minus in the sequence has the value of $\ge 0$ . Since $x \ge \dfrac{1}{2}\left(1+\sqrt{1+4x}\right)$ for all $x \ge 2$ , regardless of the number of positive $x$ 's to the next minus, the radical containing that next minus will always be positive and yield a real value. Thus, all such infinite nested radicals are real for $x \ge 2$ .

In all this, we work from the most embedded radicals outward to the "largest" radical, the one that is not embedded at all.

Here plotted are some selected infinite nested radicals that have a small number of minuses to give one an idea. The plot that has real values for $0 \le x \le 2$ is the case where there are no minuses.

Note: If one tries to consider a case where there are "minuses at the end of of all pluses", that implies that there is an end to all pluses, when there is not. Hence, the most embedded radical which contains no minuses has the value $\dfrac{1}{2}\left(1+\sqrt{1+4x}\right)$ , and we then only need to consider a finite nested radical with all the finite number of minuses in it.