Real value of equation $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$

Question

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.

Answer 1

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, $\frac{1+\sqrt{5}}{2}\approx1.613\ldots$.

Answer 2

$(x-1/x)=(x-(1-1/x)^{1/2}) ^{2} \leftrightarrow x=x^{2}+1-2x(1-1/x)^{1/2} \leftrightarrow (x^{2}-x+1)^{2}=4x^{2}(1-1/x) $

Answer 3

We may assume $x>0$. From $\sqrt{x^2-1}+\sqrt{x-1}=x\sqrt{x}$, subtract $\sqrt{x-1}$ from both side and squaring gives us $$x^2-1=x^3+x-1-2x\sqrt{x^2-x}$$ Simplifying, $$(x^2-x)-2\sqrt{x^2-x}+1=0$$ Can you go on from here?