Question:
If $$x = \frac{4\sqrt{2}}{\sqrt{2}+1}$$ Then find value of, $$\frac{1}{\sqrt{2}}*(\frac{x+2}{x-2}+\frac{x+2\sqrt{2}}{x - 2\sqrt{2}})$$
My approach:
I rationalized the value of $x$ to be $8-4\sqrt{2}$, then substituted values to get:
$$\frac{1}{\sqrt{2}}* (\frac{10 - 4\sqrt{2}}{6 - 4\sqrt{2}}+\frac{8-6\sqrt{2}}{8-2\sqrt{2}})$$
and solved until I got:
$$\frac{24-15\sqrt{2}}{8\sqrt{2}-11}$$
But this doesn't seem to please the options.
Can anyone please guide me on how to approach this problem (not all the steps, that would be huge, but the beginning steps or any hints).
Thanks a lot.
$$\frac x2=\frac{2\sqrt2}{\sqrt2+1}$$
Applying Componendo & Dividendo, $$\frac{x+2}{x-2}=\frac{2\sqrt2+(\sqrt2+1)}{2\sqrt2-(\sqrt2+1)}=\frac{3\sqrt2+1}{\sqrt2-1}=\frac{(3\sqrt2+1)(\sqrt2+1)}{2-1}$$
$$=7+4\sqrt2$$
Similarly, $$\frac x{2\sqrt2}=\frac2{\sqrt2+1}$$
Apply Componendo & Dividendo again