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: $$\dfrac a1=\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$$
calling for Componendo and Dividendo
$$\dfrac{a+1}{a-1}=\dfrac{\sqrt{x+2}}{\sqrt{x-2}}$$
Squaring we get  $$\dfrac{a^2+1+2a}{a^2+1-2a}=\dfrac{x+2}{x-2}$$
Again apply componendo and dividendo,  $$\dfrac{a^2+1}{2a}=\dfrac x2$$
Now simplify
A: Hint. Note after the rationalization you should get $a=\frac{x+ \sqrt {x^2-4}}{2}$ which is a solution of the quadratic equation $z^2-xz+1=0$.
A: You should check your rationalization of $a$ again. I believe you are missing a factor of $\frac{1}{2}$.
Additionally, you could find the answer by choosing some easily computable value of $x$, say $x = 2$, so $a = 1$ and $a^2 - ax = -1$.
A: Rationalizing $a$, you should get $$\frac{x+\sqrt{x^2-4}}{2}$$
A: HINT: write $a$ in the form $$a=\frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}=\frac{(\sqrt{x+2}+\sqrt{x-2})^2}{4}$$
A: Doing rationalization you get:
$$a=\frac{x+\sqrt{x^2-4}}{2}\to a-x=\frac{\sqrt{x^2-4}-x}{2}$$
so,
$$a(a-x)=\frac{(\sqrt{x^2-4})^2-x^2}{4}=-1$$
A: $$a=\frac{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)\left(\sqrt{x+2}+\sqrt{x-2}\right)}=\frac{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}{4},$$ 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
$$a^2-ax=a^2-a^2-1=-1$$
