If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to 
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$  then
  the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)

My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{x^2-4)^2}=2x^2-4+2x\sqrt{x^2-4}$
Now,
$a^2-ax=2x^2-4+2x\sqrt{x^2-4}-x^2-x\sqrt{x^2-4}=x^2+x\sqrt{x^2-4}-4=xa-4$
Why am I not getting the intended value?

$$a=\frac{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)\left(\sqrt{x+2}+\sqrt{x-2}\right)}=\frac{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}{4},$$ which gives $\sqrt{x+2}+\sqrt{x-2}=2\sqrt{a}$ and from here $\sqrt{x+2}-\sqrt{x-2}=\frac{2}{\sqrt{a}}.$
After summing of  last two equalities we obtain $\sqrt{x+2}=\sqrt{a}+\frac{1}{\sqrt{a}}$ or
$x=a+\frac{1}{a}$,
which gives the answer: $-1$ because
$$a^2-ax=a^2-a^2-1=-1$$
