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

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$