Simplify, $\sqrt {7 - \sqrt {7 + \sqrt {7 - \sqrt {7 + .............\infty {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} terms} } } } $

Let, $y = \sqrt {7 - \sqrt {7 + \sqrt {7 - \sqrt {7 + .............} } } } $

$\therefore y = \sqrt {7 - \sqrt {7 + y} } $

Squaring, ${y^2} = 7 - \sqrt {7 + y} $

Or, $7 - {y^2} = \sqrt {7 + y} $

Squaring, $49 - 14{y^2} + {y^4} = 7 + y$

$\therefore {y^4} - 14{y^2} - y + 42 = 0$

Using factor theorem,

${y^3}(y - 2) + 2{y^2}(y - 2) - 10y(y - 2) - 21(y - 2) = 0$

$ \Rightarrow (y - 2)({y^3} + 2{y^2} - 10y - 21) = 0$

y=2 is one possible solution. For other solutions, consider ${y^3} + 2{y^2} - 10y - 21 = 0$

Using factor theorem,

${y^2}(y + 3) - y(y + 3) - 7(y + 3) = 0$

$ \Rightarrow (y + 3)({y^2} - y - 7) = 0$

y=-3 could be one possible solution but is rejected since $y = \sqrt {7 - \sqrt {7 + \sqrt {7 - \sqrt {7 + .............} } } } > 0$

For other solutions, consider ${y^2} - y - 7 = 0$

$y = \frac{{1 \pm \sqrt {1 + 28} }}{2} = \frac{{1 \pm \sqrt {29} }}{2}$

Since y > 0, the only other possible solution can be $\frac{{1 + \sqrt {29} }}{2}$

Since squaring is involved, both possible solutions i.e. 2 and $\frac{{1 + \sqrt {29} }}{2}$ need to be checked.

Using $y = \sqrt {7 - \sqrt {7 + y} } $ to check.

$\sqrt {7 - \sqrt {7 + 2} }  = \sqrt {7 - \sqrt 9 }  = \sqrt {7 - 3}  = 2 = y$

That means 2 is indeed a solution.

Now, $\frac{{1 + \sqrt {29} }}{2} \approx \frac{{1 + 5.4}}{2} = 3.2$

$\sqrt {7 - \sqrt {7 + y} }  \approx \sqrt {7 - \sqrt {10.2} }  \approx \sqrt {7 - 3.2}  = \sqrt {3.8}  < 3.2$

This means $\frac{{1 + \sqrt {29} }}{2}$ is NOT a solution.

So, $\sqrt {7 - \sqrt {7 + \sqrt {7 - \sqrt {7 + .............\infty {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} terms} } } } =2$