Solve the Mystery: \[x + \frac{1}{x}\]?

[lfdahl]

If

\[x^{x-\sqrt{x}} = \sqrt{x}+1\]- then what is the value of \[x + \frac{1}{x}?\]

 

[Opalg]

If

\[x^{x-\sqrt{x}} = \sqrt{x}+1\]- then what is the value of \[x + \frac{1}{x}?\]

Partial answer:
[sp]
Suppose that $x = \sqrt x + 1$. Then $x - \sqrt x = 1$, and the given equation becomes $x^1 = x$, which is obviously true. But if $x = \sqrt x + 1$ then $\sqrt x = x-1$ and so (after squaring both sides) $x^2 - 3x + 1 = 0$. Therefore $\boxed{x + \dfrac1x = 3}$.

The positive solution of $x^2 - 3x + 1 = 0$ is $x = \frac12(3 + \sqrt5) \approx 2.628$ (which is in fact the square of the golden ratio). But a graph of the function $y = x^{x-\sqrt{x}} - \sqrt{x}-1$ shows that this function has two zeros. One of them is $x \approx 2.628$, as above. But there is another zero, $x \approx 0.215732$, and for that zero $\boxed{x + \dfrac1x \approx 4.85111}.$ I have no idea how to arrive at that solution, or whether there is an exact expression for it.

[/sp]

 

[lfdahl]

Partial answer:
[sp]
Suppose that $x = \sqrt x + 1$. Then $x - \sqrt x = 1$, and the given equation becomes $x^1 = x$, which is obviously true. But if $x = \sqrt x + 1$ then $\sqrt x = x-1$ and so (after squaring both sides) $x^2 - 3x + 1 = 0$. Therefore $\boxed{x + \dfrac1x = 3}$.

The positive solution of $x^2 - 3x + 1 = 0$ is $x = \frac12(3 + \sqrt5) \approx 2.628$ (which is in fact the square of the golden ratio). But a graph of the function $y = x^{x-\sqrt{x}} - \sqrt{x}-1$ shows that this function has two zeros. One of them is $x \approx 2.628$, as above. But there is another zero, $x \approx 0.215732$, and for that zero $\boxed{x + \dfrac1x \approx 4.85111}.$ I have no idea how to arrive at that solution, or whether there is an exact expression for it.

[/sp]

Thankyou very much, Opalg, for your in depth analysis of the problem!

 

[Albert1]

Partial answer:
[sp]
Suppose that $x = \sqrt x + 1$. Then $x - \sqrt x = 1$, and the given equation becomes $x^1 = x$, which is obviously true. But if $x = \sqrt x + 1$ then $\sqrt x = x-1$ and so (after squaring both sides) $x^2 - 3x + 1 = 0$. Therefore $\boxed{x + \dfrac1x = 3}$.

The positive solution of $x^2 - 3x + 1 = 0$ is $x = \frac12(3 + \sqrt5) \approx 2.628$ (which is in fact the square of the golden ratio). But a graph of the function $y = x^{x-\sqrt{x}} - \sqrt{x}-1$ shows that this function has two zeros. One of them is $x \approx 2.628$, as above. But there is another zero, $x \approx 0.215732$, and for that zero $\boxed{x + \dfrac1x \approx 4.85111}.$ I have no idea how to arrive at that solution, or whether there is an exact expression for it.

[/sp]

my solution (a clue from Opalg's solution)

Spoiler

let :$x-\sqrt{x}=p---(1)$
we have :$x^p-1=x-p--(2)$
square (1):$x^2-2px+p^2=x---(3)\\$
square (2):$x^{2p}-2x^p+1=x^2-2px+p^2=x---(4)$
compare (3) and (4)
if $p=1$
so $x^2-3x+1=0$
$x=\dfrac {3+\sqrt 5}{2}$
$x+\dfrac{1}{x}=3 $
if $p=-1/4$
from 2 :$x\approx 0.2157$( intersection of $y=x+\dfrac {1}{4} ,\,\, and \,\, y=x^{-1/4}-1$
and $x+\dfrac {1}{x}\approx 4.85$

 

[kaliprasad]

my solution:

Spoiler

let :$x-\sqrt{x}=p---(1)$
we have :$x^p-1=x-p--(2)$
square (1):$x^2-2px+p^2=x---(3)\\$
square (2):$x^{2p}-2x^p+1=x^2-2px+p^2=x---(4)$
compare (3) and (4)
if $p=1$
so $x^2-3x+1=0$
$x=\dfrac {3+\sqrt 5}{2}$
$x+\dfrac{1}{x}=3 $
if $p=-1/4$
from 2 :$x\approx 0.2157$( intersection of $y=x+\dfrac {1}{4} ,\,\, and \,\, y=x^{-1/4}-1$
and $x+\dfrac {1}{x}\approx 4.85$

how is

Spoiler

$p= -\frac{1}{4}$

 

[Albert1]

how is

Spoiler

$p= -\frac{1}{4}$

Spoiler

$p=\dfrac {-1}{4}$ is also a possible solution of (1)
and put $p=\dfrac{-1}{4}$ to (2) to find solution of $x$
the intersections of (4):$ x^{2p}-2x^p+1=x^2-2px+p^2=x$
are $p=1,\,\ p=\dfrac {-1}{4}$
in fact we can find another $x$ from $x^2-2px+p^2=x^{2p}-2x^p+1$
or $(x-p=x^p-1)$
(let $p=\dfrac {-1}{4}$)

 

[lfdahl]

my solution (a clue from Opalg's solution)

Spoiler

let :$x-\sqrt{x}=p---(1)$
we have :$x^p-1=x-p--(2)$
square (1):$x^2-2px+p^2=x---(3)\\$
square (2):$x^{2p}-2x^p+1=x^2-2px+p^2=x---(4)$
compare (3) and (4)
if $p=1$
so $x^2-3x+1=0$
$x=\dfrac {3+\sqrt 5}{2}$
$x+\dfrac{1}{x}=3 $
if $p=-1/4$
from 2 :$x\approx 0.2157$( intersection of $y=x+\dfrac {1}{4} ,\,\, and \,\, y=x^{-1/4}-1$
and $x+\dfrac {1}{x}\approx 4.85$

Brilliant, Albert! Thankyou for your contribution!