<s>[INST] How to solve $x^2 +\left(\frac{x}{x-1}\right)^2 =8$? I tried to solve this question but it turns into a 4th degree equation and I could only get one solution for this equation,i.e., 2.
It is to be evaluated for solutions. Thanks.
 [/INST] $$x^2 +\frac{x^2}{(x-1)^2} = 8$$
$$x^2 +\frac{x^2}{(x^2-2x + 1)} = 8$$
$$x^2 = (8 - x^2)(x^2-2x + 1)$$
$$x^2 = (8 - x^2)x^2-(8 - x^2)2x + (8 - x^2)$$
$$x^2 = 8x^2 - x^4 -16x + 2x^3 - x^2 + 8$$
$$0 =  - x^4 -16x + 2x^3 + 6x^2 + 8  $$
$$(x-2)^2(x^2+2x-2)= 0$$
$$x= 2$$
$$x= \sqrt{3}-1 $$
$$x= -1 - \sqrt{3}$$
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