<s>[INST] How to find $\sqrt{1+{4\over x}+{4\over x^2} }$? If   $$abx^2 = (a-b)^2(x+1)$$ then what is $$\sqrt{1+{4\over x}+{4\over x^2} }$$ 

(A) $a+b \over a-b$  (B)$a-b\over a+b$ (C) $a^2+ab$ (d) None

EDIT: What I've done is this:
$$abx^2=(a^2+b^2-2ab)(x+1)$$
=> $$abx^2 - (a^2+b^2+2ab)x - (a-b)^2=0$$
=> $$abx^2 - (a-b)^2x - (a-b)^2=0$$
but this is getting me nowhere.
 [/INST] this is $({1+\frac{2}{x}})^2$ which is $1+\frac{2}{x}$ according to your question. Then after what You have done try finding the roots by using $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$.
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