The double square roots can be removed by $8+2\sqrt7 = (\sqrt7+1)^2$ and $$4-\sqrt7 = \frac{1}{4}(16-4\sqrt7) = \frac{1}{4}(14+2-2\sqrt2\cdot\sqrt14) = \frac{1}{4}(\sqrt14-\sqrt2)^2$$The expression simplifies to $\frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2}$. This problem was given to me as an MCQ, the four options were:
A) $1$,
B) $2$,
C)$32\sqrt3$
D) $36\sqrt3$.
I am not getting any of these options, $$\begin{eqnarray*}\frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2} &=& \frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2} \cdot \frac{\sqrt7+1+\sqrt2}{\sqrt7+1+\sqrt2} \\ &=& \frac{1}{2}\cdot \frac{6\sqrt2+2\sqrt7-2}{6+2\sqrt7} \\ &=& \frac{1}{4}(9\sqrt2-3\sqrt14+4\sqrt7-10) \end{eqnarray*}$$ Is my calculation incorrect ?
Let $\sqrt{8+2\sqrt{7}}=\sqrt{x}+\sqrt{y} \implies x+y+2\sqrt{xy}, \implies x+y=8, xy=7 \implies x=1,y=7$ and let $\sqrt{4-\sqrt{7}}=\sqrt{p}-\sqrt{q} \implies p+q=4, pq=7/4 \implies p=7/2, q=1/2.$ Then $$F=\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}-\sqrt{2}}= \frac{\sqrt{7/2}-\sqrt{1/2}}{1+\sqrt{7}-\sqrt{2}}$$
Something is amiss!