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Find all real numbers of $x$ such that $$x=\left(x-\frac 1x\right)^{\frac 12}+\left(1-\frac 1x\right)^{\frac 12}$$

My Attempt: Square both sides to get$$x^2=x-\frac 1x+1-\frac 1x+2\sqrt{\left(x-\frac 1x\right)\left(1-\frac 1x\right)}$$ Then move all the terms to the left side except for the square root and then square the equation. But I predict the terms to get really ugly really fast.

So I'm wondering if there is an elegant way of obtaining the solution without going through much of a hassle.

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  • $\begingroup$ First put $x+\sqrt{x-1/x}$ on one side, then square. After some cancelling and another squaring you should get $x^4 - 2x^3 - x^2 + 2x + 1 = (1+x-x^2)^2 = 0$. $\endgroup$ – Ivica Smolić Oct 8 '16 at 15:06
  • $\begingroup$ I ment to write $x - \sqrt{x - 1/x}$ $\endgroup$ – Ivica Smolić Oct 8 '16 at 15:26
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Multiply the original equation $$ (x-x^{-1})^{1/2}+(1-x^{-1})^{1/2}=x\tag{1} $$ by $(x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}$ to get $$ x-1=x((x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}) $$ That is $$ (x-x^{-1})^{1/2}-(1-x^{-1})^{1/2}=1-x^{-1}\tag{2} $$ Denote $a=(x-x^{-1})^{1/2}$ and $b=(1-x^{-1})^{1/2}$. Then $(1)$ and $(2)$ can be rewritten as $$ a+b=x\tag{3} $$ $$ a-b=b^2\tag{4} $$ From the very definition of $a$ and $b$ we have $a^2-b^2=x-1$. With $(3)$ we get $$ a^2-b^2=a+b-1\tag{5} $$ Subtracting $(4)$ from $(5)$ we obtain $a^2-a=a-1$. Hence $a=1$. The rest is clear.

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The RHS of the given equation is $\geq0$, hence we need only consider solutions $x\geq0$. In this range the square roots on the RHS are only defined when $x\geq1$. We therefore may restrict our search to $x\geq1$.

Put $$x-{1\over x}=:u^2,\qquad 1-{1\over x}=:v^2$$ with $u\geq0$, $v\geq0$. Then $u+v=x$ and $$u-v={u^2-v^2\over u+v}={x-1\over x}=1-{1\over x}\ .$$ This leads to $2v=x+{1\over x}-1$ and therefore $$\left(x+{1\over x}-1\right)^2=4v^2=4-{4\over x}\ .$$ After multiplying by $x^2$ and collecting terms we then obtain $$0=x^4-2x^3-x^2+2x+1=\bigl(x^2-x-1\bigr)^2\ .$$ The equation $x^2-x-1=0$ has the solutions $\xi:={\sqrt{5}+1\over2}\doteq1.618$ and $-{1\over\xi}<0$. It is easy to check that $\xi$ is indeed a solution (and therewith the only solution) of the original problem:

Note that $\xi-1={1\over\xi}$. Therefore $\xi-{1\over\xi}=1$, and $$1-{1\over\xi}={\xi-1\over\xi}={1\over\xi^2}\ .$$ It follows that $$\sqrt{\xi-{1\over\xi}}+\sqrt{1-{1\over\xi}}=1+{1\over\xi}=\xi\ .\qquad\square$$

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