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I was tackling through an olympiad practice book when I saw one of these problems:

If $x = 5 + 2\sqrt6$, evaluate $\Large{x \ - \ 1 \over\sqrt{x}}$?

The answer written is $2\sqrt2$, but I can't figure my way out through the manipulations. I just know that I have the following:$${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$

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My next step would be multiplying by the conjugate: $\frac {\sqrt{ 5-2\sqrt 6}}{\sqrt {5-2\sqrt 6}}$. – Ross Millikan Dec 3 '12 at 14:54
There is no need for a new tag, certainly not one with such a backstabbing, Machiavellian name like [manipulation]. Also, please avoid using $$ environment (or generally display equations) in the title. – Asaf Karagila Dec 3 '12 at 14:57
@AsafKaragila: Yes, I thought for once that the display became too large. – Parth Kohli Dec 3 '12 at 15:01
up vote 16 down vote accepted

Hint $\ \ $ Squaring it, we get $\ \rm\dfrac{40+16\sqrt{6}}{5+2\sqrt{6}}\, =\, 8$

Remark $\ \ $ Generally $\rm\ \ \dfrac{ n\!-\!1 + \sqrt{n^2\!-\!1}}{ \sqrt{ n\ +\ \sqrt{n^2\!-\!1}}}\, =\, \sqrt{2(n\!-\!1)}\ $ for $\rm\: n\ge 1,\:$ with analogous proof.

For another example note $\, \dfrac{\quad\ 3 + \sqrt{11}}{\sqrt{10+ \sqrt{99}}}\, =\, \sqrt{2},\ $ by $\rm\:n = 10\:$ above.

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Isn't this how you show 5 equals 4? – rubenvb Dec 3 '12 at 15:44
@rubenvb $\rm\ x^2 = y^2\Rightarrow\: x = y\:$ for positive reals $\rm\:x,y\ \ $ – Bill Dubuque Dec 3 '12 at 15:50

Start by multiplying the numerator and denominator of $${4+ 2\sqrt6} \over {\sqrt{5 + 2\sqrt6}}$$ by $\sqrt{5 - 2\sqrt{6}}$.

Then the denominator evaluates very nicely to 1. Then it's much a matter of simplifying the numerator.

Simpler yet, square the expression, simplify, and then take the square root of the result.

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Thanks, I get the idea. You're trying to rationalize the denominator. :) – Parth Kohli Dec 3 '12 at 14:59

When you know the answer, it is often easier...

Indeed, begin with $(x-1)/\sqrt{x} = 2\sqrt{2}$

It is equivalent to $x-1 = 2\sqrt{2x}$

Equivalent to $(x-1)^2 = 8x$

Equivalent to $x^2-2x+1 = 8x$

Equivalent to $x^2-10x+1 = 0$

Equivalent to $(x-a)(x-b) = 0$ where $a = 5+2\sqrt{6}$ and $b = 5-2\sqrt{6}$

Equivalent to {$x=a$ or $x=b$}

Of course, since these are equivalences, this suffices to prove the fact. However, if you want to write it in a more straightforward way, it suffices to go back from the end to the beginning:

Let $x = 5+2\sqrt{6}$. Then $(x-(5+2\sqrt{6})(x-(5-2\sqrt{6}))=0$. But this product can be expanded as $x^2-10x+1$. So, we have $x^2+1 = 10x$. At this point, you can generate various identities. For instance, you could conclude $x^2+2x+1 = 12x$. Therefore $(x+1)^2=12x$ and $x+1 = 2\sqrt{3x}$ and so $(x+1)/\sqrt{x} = 2\sqrt{3}$...

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Hint: Note that, since $2+3=5$ and $2\cdot 3=6$, we have that $x=5+2\sqrt6=(\sqrt2+\sqrt3)^2$. And that the numerator $=2\sqrt2\cdot (\sqrt2+\sqrt3)$.

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yes, numerator, thx – Berci Dec 5 '12 at 0:16

$$\frac{x-2}{\sqrt x}=\frac{4+ 2\sqrt6}{\sqrt{5 + 2\sqrt6}}=\frac{4+ 2\sqrt6}{\sqrt{5 + 2\sqrt6}}\cdot\frac{\sqrt{5-2\sqrt 6}}{\sqrt{5-2\sqrt 6}}=\left(4+2\sqrt 6\right)\sqrt{5-2\sqrt 6}\Longrightarrow$$

$$(x-2)^2=x(40+16\sqrt 6)(5-2\sqrt 6)=x(8)\Longrightarrow $$


Take it from here

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The square of $\sqrt{x} - {1 \over \sqrt{x}}$ is $x - 2 + {1 \over x}$. In this case $x = 5 + 2\sqrt{6}$, whose reciprocal is seen to be $5 - 2\sqrt{6}$ by rationalizing the denominator. So $$x - 2 + {1 \over x} = (5 + 2\sqrt{6}) - 2 + (5 - 2\sqrt{6})$$ $$= 8$$ So your answer is $\sqrt{8} = 2\sqrt{2}$. (You take the positive square root since $x > 1$).

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