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.
 A: We have $$x^2 +\frac {x^2}{(x-1)^2} =8$$ $$\Rightarrow x^2(x^2-2x+2) = 8 (x^2-2x+1) $$ 
$$\Rightarrow x^4-2x^3-6x^2+16x-8 =0$$ 
$$\Rightarrow (x-2)(x^3-6x+4)=0$$
$$\Rightarrow  (x-2)(x-2)(x^2+2x-2)=0$$
Hope you can take it from here by using the quadratic formula to factor the last bit.
A: Note that multiplying $(x-1)^2$ on each side gives us that
$$x^2(x-1)^2+x^2=8(x-1)^2$$
Note that this is $$x^2(x-1)^2+x^2+(x-1)^2=9(x-1)^2$$
However, note that this implies $$(x^2-x+1)^2=9(x-1)^2$$
As seen here. So $$x^2-x+1= \pm 3(x-1)$$
I think you can continue from here. 
A: writing your equation in the form
$$x^2+\left(\frac{x}{x-1}\right)^2-8=0$$
and factorizing we obtain
$$\left( {x}^{2}+2\,x-2 \right)  \left( x-2 \right) ^{2}=0$$
can you finish now?
p.s.: it must be $$x\ne 1$$
A: $$x^2 +\frac{x^2}{(x-1)^2} = 8$$
$$x^2 +\frac{x^2}{(x^2-2x + 1)} = 8$$
$$x^2 = (8 - x^2)(x^2-2x + 1)$$
$$x^2 = (8 - x^2)x^2-(8 - x^2)2x + (8 - x^2)$$
$$x^2 = 8x^2 - x^4 -16x + 2x^3 - x^2 + 8$$
$$0 =  - x^4 -16x + 2x^3 + 6x^2 + 8  $$
$$(x-2)^2(x^2+2x-2)= 0$$
$$x= 2$$
$$x= \sqrt{3}-1 $$
$$x= -1 - \sqrt{3}$$
