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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

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