Find all $x \in \mathbb{R}$ such that $1+\sqrt{1-\frac{1}{x}}+\sqrt{x-\frac{1}{x}}=x^2$.

Question

This problem was posed by a friend of mine. He's got it from a social media post. We don't know how to solve it, but there are similar types of questions in this website, whose techniques I've tried to apply but to no avail.

I'm trying to find some sort of manipulation that allows us to find the real roots of the equation: \begin{equation*}1+\sqrt{1-\frac{1}{x}}+\sqrt{x-\frac{1}{x}}=x^2\end{equation*} \begin{align*}\text{Where } x \in \mathbb{R} \end{align*}

I've already tried many things, but this is what seems cleaner: \begin{align*} a=\sqrt{x-\frac{1}{x}},& \quad b=\sqrt{1-\frac{1}{x}}\\ \implies a+b=x^2-1,&\quad a-b=\frac{1}{x+1}\\ \to a^2+b^2=\frac{x^2+x-2}{x}, &\quad a^2-b^2=x-1 \end{align*} But I couldn't get much further. This looks like a method where I could stay organized and try to spot something with more ease, but I wasn't able to do so.

I'm not interested in employing numerical methods. I know $1$ is a trivial solution and, through the usage of software, that $\varphi$ is also a solution.

Thanks in advance.

Answer 1

Subtract 1 from both sides, and then multiply both sides by $\sqrt{x}$:

$$\sqrt{x-1} + \sqrt{x^2-1} = (x^2-1) \sqrt{x}.$$

Square both sides:

$$(x-1) + 2(x-1)\sqrt{x+1} + (x^2-1) = (x^2-1)^2 x.$$

Collect terms:

$$2(x-1)\sqrt{x+1} = x^5 - 2x^3 - x^2 + 2.$$

Square both sides:

$$2(x-1)^2 (x+1) = (x^5 - 2x^3 - x^2 + 2)^2.$$

Expand, and collect terms:

$$x^{10} - 4x^8 - 2x^7 + 4x^6 + 8 x^5 + x^4-12 x^3 + 4x = 0.$$

Factor the LHS:

$$(x-1)^2 x (x^2-x-1) (x^5+3x^4+3x^3-4x-4) = 0.$$

This immediately shows that $x=0$, $x=1$ and $x=(1\pm\sqrt{5})/2$ are possible roots. Plugging in shows that $x=0$ and $x=(1-\sqrt{5})/2$ are spurious, but $x=1$ and $x=(1+\sqrt{5})/2$ are actual roots. $x^5+3x^4+3x^3-4x-4$ has one real root but it does not satisfy the original equation. Therefore, there are no other solutions to the original equation.

As the comments indicate, there are more elegant and simpler ways to get to a solution, but I wanted to show a brute-force method that requires little or no cleverness and just applies straightforward algebra by "following your nose".

Answer 2

We are given $$f(x):=1+\sqrt{1-\frac1x}+\sqrt{x-\frac1x}=x^2.$$ For the expression for $f(x)$ to be meaningful, we must have $x>-1$ and $x\neq0$. Also clearly $f(x)\geqslant1$. Hence also $x^2\geqslant1$. Therefore $x\geqslant1$.

Obviously $x=1$ is a solution. We will show that, for $x>1$, the gradient of the curve $y=f(x)$ is initially infinitely steep, and so exceeds the gradient of $y=x^2$, while it declines thereafter, so that the curves $y=f(x)$ and $y=x^2$ intersect just once beyond $x=1$ (because the gradient of the latter curve is increasing without bound). We have $$f'(x)=\frac1{2x^2}\left(1-\frac1x\right)^{\!-1/2}+\frac12\left(1+\frac1{x^2}\right)\!\left(x-\frac1x\right)^{\!-1/2}.$$ Since $1-1/x$ and $x-1/x$ approach $0$ as $x$ approaches $1$, it is clear that $f'(x)\to\infty$ as $x\to1+$. Now $$f''(x)=-\frac1{4x^4}\left(1-\frac1x\right)^{\!-3/2}-\frac1{x^3}\left(1-\frac1x\right)^{\!-1/2}$$

$$\qquad\qquad\qquad\qquad\qquad-\frac14\left(1+\frac1{x^2}\right)^{\!2}\left(x-\frac1x\right)^{\!-3/2}-\frac1{x^3}\left(x-\frac1x\right)^{\!-1/2}.$$ As the sum of four negative terms, this is negative, and so $f'(x)$ is decreasing, for $x>1$. It follows that our given equation has just two solutions: $x=1$ and a further value for $x$ between $1$ and $2$. By trying a few simple values, we can see that $1.5$ is too low, and $1.6$ is pretty close but still a bit low.

At this point, we take a break from the question and consider the setting. The questioner can't expect us to calculate an irrational solution to an ugly high-order polynomial equation. There must be a “simple” answer which magically simplifies the equation. We observe that the golden ratio $\phi$ satisfies simplifying identities, such as $$\frac1\phi=\phi-1,\qquad\phi^2=\phi+1,\qquad\sqrt{2-\phi}=\phi-1.$$ We saw that the second solution had to be a bit bigger than $1.6$; so let us give $\phi$ a try. Indeed, using the simplifying identities, it works! So $\phi$ must be the unique solution other than $1$.