<s>[INST] Real value of equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$ Find the real value of x in the equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$
I tried to square the whole term and after expansion not getting the result.
 [/INST] Note for the expression $(1-1/x)^{1/2}$ to have meaning, $1$ must be greater than or equal to $1/x$. So $x\geq1$.
Squaring both sides, $$x-\frac{1}{x}+2\left(x-1-\frac{1}{x}+\frac{1}{x^2}\right)^{1/2}+1-\frac{1}{x}=x^2$$
Rearranging:
$$2\left(x-1-\frac{1}{x}+\frac{1}{x^2}\right)^{1/2}=x^2-x-1+\frac{2}{x}$$
Squaring again:
$$4\left(x-1-\frac{1}{x}+\frac{1}{x^2}\right)=x^4-2x^3-x^2+6x-3-\frac{4}{x}+\frac{4}{x^2}$$
$$0=x^4-2x^3-x^2+2x+1$$
$$0=\left(x^2-x-1\right)^2$$
$$0=x^2-x-1$$
There is one solution greater than $1$, the large Golden Ratio, $\frac{1+\sqrt{5}}{2}\approx1.613\ldots$.
</s>