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.

  • $\begingroup$ 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

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.


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 $$\sqrt{\xi-{1\over\xi}}+\sqrt{1-{1\over\xi}}=1+{1\over\xi}=\xi\ .\qquad\square$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.