Show that $\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$

Question

Show that $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}.$$

My attempt: I could find a way to develop the LHS by un-nesting the double radical, figuring out that $$\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}\ \ (1)$$ By substituting (1) in the LHS of the original expression, it is straightforward to show that it is equal to the RHS.

But my problem is on how to show without the un-nesting, by another approach. I've used a standard approach of multiplying both terms by $\sqrt{2}+\sqrt{2+\sqrt{3}}$, leading to

$$\frac{4+\sqrt{3}+2\sqrt{4+2\sqrt{3}}}{-\sqrt{3}}\Leftrightarrow \frac{(4+\sqrt{3}+2\sqrt{4+2\sqrt{3}})\sqrt{3}}{-3}.$$

But I was not able to show that this last expression is equal to RHS by this approach.

Hints and answers not using my first un-nesting approach (if possible) will be appreciated. Sorry if this is a duplicate.

Answer 1

The simple, "natural" solution is using direct radical denesting, as noted by the OP. But any algebraic solution will likely involve some form of "denesting" eventually, as noted by Misha Lavrov in a comment. For example, the following is a rather convoluted, very "artificial" solution using polynomial resultants.

Let $\,t_{1,2} = \sqrt{2} \pm \sqrt{2+\sqrt{3}}\,$, then $\,t_1+t_2=2 \sqrt{2}\,$ and $\,t_1t_2=-\sqrt{3}\,$, so $\,t_{1,2}\,$ are the roots of:

$$ t^2 - 2 \sqrt{2} \,t - \sqrt{3} = 0 \tag{1} $$

Let $\,u = t_1 / t_2\,$, then $\,t_2\,$ and $\,u t_2 = t_1\,$ both satisfy $\,(1)\,$, so there is a common root between $\,(1)$ and:

$$ u^2 t^2 - 2 \sqrt{2} \,ut - \sqrt{3} = 0 \tag{2} $$

Eliminating $\,t\,$ between $\,(1)\,$ and $\,(2)\,$ by means of resultants gives a quartic in $\,u\,$, namely $\,\operatorname{res}_{t}(\,t^2 - 2 \sqrt{2} \,t - \sqrt{3}\,,\; u^2 t^2 - 2 \sqrt{2} \,ut - \sqrt{3}\,) = 0\,$. Obviously $\,u=1\,$ must be a root, and in fact a double root, which helps with factoring it as:

$$ 3 u^4 + 8 \sqrt{3} u^3 - 16 \sqrt{3} u^2 - 6 u^2 + 8 \sqrt{3} u + 3 = (u-1)^2\big(3 u^2 + 2\,(3 + 4 \sqrt{3})\, u + 3\big) \tag{3} $$

Since we are looking for $\,u \ne 1$, our $\,u\,$ must be a root of the latter quadratic factor. It can be shown that it's the smaller of the two roots, so $\,u = \frac{1}{3}\left(-(3+4\sqrt{3}) - 2 \sqrt{6\,(2+\sqrt{3})}\right)\,$. However, simplifying it to $\,-3 - 2 \sqrt{3}\,$ requires the same radical denesting that would have easily solved the problem in the first place.

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[ EDIT ]   If, however, the end result is known in advance, then it is fairly easy to prove it directly without any explicit denesting. By rearranging the terms and isolating the nested radical:

$$ \begin{align} \frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3} \quad\iff\quad \sqrt{2+\sqrt{3}} &= \sqrt{2}\,\frac{4+2 \sqrt{3}}{2+ 2 \sqrt{3}}\\ &= \sqrt{2}\,\frac{2 + \sqrt{3}}{1 + \sqrt{3}} \cdot \color{red}{\frac{\sqrt{3}-1}{\sqrt{3}-1}} \\ &= \sqrt{2}\,\frac{1+\sqrt{3}}{2} \end{align} $$

Since both sides of the latter equality are positive, it is enough to show that their squares are equal:

$$ \left(\sqrt{2+\sqrt{3}}\right)^2 = \left(\sqrt{2}\,\frac{1+\sqrt{3}}{2}\right)^2 \quad\iff\quad 2 + \sqrt{3} = 2 \cdot \frac{4 + 2 \sqrt{3}}{4} $$

Answer 2

You need to denest $\sqrt{2+\sqrt{3}}$ at some point, but this can be deferred to the last step. Let $s=\sqrt{2}$ and $r=\sqrt{3}$. Then \begin{align} \frac{s+\sqrt{s^2+r}}{s-\sqrt{s^2+r}} +3+2r &=\frac{1+\sqrt{1+r/2}}{1-\sqrt{1+r/2}} +3+2r\\ &=\frac{1+\sqrt{1+r/2}}{1-\sqrt{1+r/2}}\times\frac{1+\sqrt{1+r/2}}{1+\sqrt{1+r/2}} +3+2r\\ &=\frac{2+r/2+2\sqrt{1+r/2}}{-r/2}+3+2r\\ &=\frac{-(r+1)+\sqrt{4+2r}}{-r/2} \end{align} and it remains to show that $\sqrt{4+2r}=r+1$.

Answer 3

we have to show, $$\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}=-3-2\sqrt{3}$$ Let,$2+\sqrt{3}=p$

$$R.H.S.= -3-2\sqrt{3}$$ $$=-2(\frac{3}{2}+\sqrt{3})$$ $$=-2\bigg(\frac{3}{2}+\frac{1}{2}-\frac{1}{2}+\sqrt{3}\bigg)$$ $$=-2(2+\sqrt{3}-\frac{1}{2})$$ $$=-2(p-\frac{1}{2})$$ $$=1-2p$$ $$=\dfrac{(2-p)(1-2p)}{2-p}$$ $$=\dfrac{2-5p+2p^2}{2-p}...........(1)$$

Now, $$L.H.S=\dfrac{\sqrt{2}+\sqrt{p}}{\sqrt{2}-\sqrt{p}}$$ $$=\dfrac{(\sqrt{2}+\sqrt{p})^2}{2-p}.............(2)$$

Now,we can conclude that if the given statement is true then,for $p=2+\sqrt{3}$,the numerators of part (1) and (2) will be equal.in reverse we can also say that the numerators will be same if $p=2+\sqrt{3}$.So,now my target is to solve for $p$ to get $2+\sqrt{3}$.

to be the statement true, $$(\sqrt{2}+\sqrt{p})^2=2-5p+2p^2$$ $$p^2-3p-\sqrt{2}p=0$$ $$p=0,2+\sqrt{3}$$

As we got a root $p=2+\sqrt{3}$,then the claim is true.$_{[Showed]}$

NOTE: you can also prove it by substituting the value of p in (2) and getting (1).but I think the above procedure is more simple.

Answer 4

Since $$\sqrt{2+\sqrt{3}} = \sqrt{4+2\sqrt{3}\over 2} = {1+\sqrt{3}\over \sqrt{2}}$$

\begin{eqnarray}\frac{\sqrt{2}+\sqrt{2+\sqrt{3}}}{\sqrt{2}-\sqrt{2+\sqrt{3}}}&=& \frac{\sqrt{2}+{1+\sqrt{3}\over \sqrt{2}}}{\sqrt{2}-{1+\sqrt{3}\over \sqrt{2}}}\\ &=& \frac{2+1+\sqrt{3}}{2-1-\sqrt{3}}\\ &=& \frac{3+\sqrt{3}}{1-\sqrt{3}}=...\\ &=&-3-2\sqrt{3} \end{eqnarray}