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?

Hint. Note after the rationalization you should get $a=\frac{x+ \sqrt {x^2-4}}{2}$ which is a solution of the quadratic equation $z^2-xz+1=0$.
