How to find $\sqrt{1+{4\over x}+{4\over x^2} }$?

Question

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.

Answer 1

$\textbf{Shortest Solution}$:

$$\frac{ab}{(a-b)^2}=\frac{(x+1)}{x^2}$$ $$\frac{ab}{(a-b)^2}=\frac{1}{x^2}+\frac{1}{x}$$

Multiplying $4$ both sides,

$$\frac{4ab}{(a-b)^2}=\frac{4}{x^2}+\frac{4}{x}$$

Adding $1$ both sides,

$$1+\frac{4ab}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$ $$\frac{(a-b)^2+4ab}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$ $$\frac{(a+b)^2}{(a-b)^2}=1+\frac{4}{x^2}+\frac{4}{x}$$

Taking Square root both sides,

$$\sqrt{\frac{(a+b)^2}{(a-b)^2}}=\sqrt{1+\frac{4}{x^2}+\frac{4}{x}}$$ $$\frac{(a+b)}{(a-b)}=\sqrt{1+\frac{4}{x^2}+\frac{4}{x}}$$

Answer 2

Hint : $$\sqrt{1+\frac{4}{x}+\frac{4}{x^2}}=\sqrt{\frac{1}{x^2}(x+2)^2}$$

Answer 3

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 $x=\frac{-b+\sqrt{b^2-4ac}}{2a}$.

Answer 4

Ans. is $\frac{a+b}{a-b}$

As @Tattwamasi Amrutam said find the value of $x$ from

$$abx^2-(a-b)^2x-(a-b)^2=0$$then

$$x=\frac{(a-b)^2 \pm \sqrt{(a-b)^4+4ab(a-b)^2}}{2ab}$$

$$x=\frac{(a-b)^2\pm \sqrt{a^4+b^4-4a^3b-4ab^3+6a^2b^2+4a^3b+4ab^3-8a^2b^2}}{2ab}$$

(I got expansion of $(a-b)^4$ using pascal's triangle)

After the simplification,

$$x=\frac{2a^2-2ab}{2ab}$$

and there is another root that u can find.

For this root only i got

$$\sqrt{1+\frac{4}{x}+\frac{4}{x^2}}=\sqrt{\bigg(1+\frac{2}{x}\bigg)^2}$$

$$1+\frac{2}{x}=1+\bigg(\frac{4ab}{2a^2-2ab}\bigg)=\frac{a+b}{a-b}$$