MHB What is the variation of Ifdahl's solution to finding the value(s) of $x$?

[Albert1]

$\sqrt {x+\dfrac {1}{x}+1}+\sqrt {x+\dfrac {1}{x}}=x $

find the value(s) of $x$

 

[lfdahl]

My attempt:

Spoiler

Given the equation:

\[\sqrt{x+\frac{1}{x}+1}+\sqrt{x+\frac{1}{x}} = x\: \: \:\: \: \: \: \: \: \; (1)\]

Let $y = x+\frac{1}{x}$.

Squaring the given equation:

\[\left ( \sqrt{y+1}+\sqrt{y} \right )^2 = x^2 \Rightarrow 2y + 1+2\sqrt{y^2+y}=x^2 \]

Squaring again:

\[\Rightarrow 4(y^2+y)=\left ( x^2-2y-1 \right )^2\] \[\Rightarrow 4y^2+4y = 4y^2+4y-4x^2y+x^4-2x^2+1 \] \[ \Rightarrow 4x^2y -x^4+2x^2-1=0\]

The last equation is a polynomial of 4th degree:

\[4x^2\left ( x+\frac{1}{x} \right )-x^4+2x^2-1 = 0 \] \[ \Rightarrow x^4-4x^3-2x^2-4x+1 = 0\]

This looks a lot like $(x-1)^4$ except for the midterm ($-2x^2$), so we´ll have to correct with $-8x^2$. The polynomial then has the form:\[(x-1)^4-8x^2 = 0\]

Taking the square root of the identity: $(x-1)^4=8x^2$ leads to:

\[x^2-(2+\sqrt{8})x+1 = 0\]

There are two positive roots \[x = \frac{1}{2}\left ( 2+\sqrt{8} \pm \sqrt{8+4\sqrt{8}} \right )\]

or

\[x = 1+\sqrt{2}\pm \sqrt{2+\sqrt{8}} = 1+\sqrt{2}\left ( 1 \pm\sqrt{1+\sqrt{2}} \right ) \approx \left\{\begin{matrix} 4.612\\0.217 = \frac{1}{4.612} \end{matrix}\right.\]

The roots are reciprocals, because the left hand side of $(1)$ is symmetric in $x$ and $\frac{1}{x}$, so both roots yield the same RHS, namely $\approx$ 4.612. Therefore, only the larger root is valid.

 

[Albert1]

My attempt:

Spoiler

Given the equation:

\[\sqrt{x+\frac{1}{x}+1}+\sqrt{x+\frac{1}{x}} = x\: \: \:\: \: \: \: \: \: \; (1)\]

Let $y = x+\frac{1}{x}$.

Squaring the given equation:

\[\left ( \sqrt{y+1}+\sqrt{y} \right )^2 = x^2 \Rightarrow 2y + 1+2\sqrt{y^2+y}=x^2 \]

Squaring again:

\[\Rightarrow 4(y^2+y)=\left ( x^2-2y-1 \right )^2\] \[\Rightarrow 4y^2+4y = 4y^2+4y-4x^2y+x^4-2x^2+1 \] \[ \Rightarrow 4x^2y -x^4+2x^2-1=0\]

The last equation is a polynomial of 4th degree:

\[4x^2\left ( x+\frac{1}{x} \right )-x^4+2x^2-1 = 0 \] \[ \Rightarrow x^4-4x^3-2x^2-4x+1 = 0\]

This looks a lot like $(x-1)^4$ except for the midterm ($-2x^2$), so we´ll have to correct with $-8x^2$. The polynomial then has the form:\[(x-1)^4-8x^2 = 0\]

Taking the square root of the identity: $(x-1)^4=8x^2$ leads to:

\[x^2-(2+\sqrt{8})x+1 = 0\]

There are two positive roots \[x = \frac{1}{2}\left ( 2+\sqrt{8} \pm \sqrt{8+4\sqrt{8}} \right )\]

or

\[x = 1+\sqrt{2}\pm \sqrt{2+\sqrt{8}} = 1+\sqrt{2}\left ( 1 \pm\sqrt{1+\sqrt{2}} \right ) \approx \left\{\begin{matrix} 4.612\\0.217 = \frac{1}{4.612} \end{matrix}\right.\]

The roots are reciprocals, because the left hand side of $(1)$ is symmetric in $x$ and $\frac{1}{x}$, so both roots yield the same RHS, namely $\approx$ 4.612. Therefore, only the larger root is valid.

Spoiler

\[x = 1+\sqrt{2}\pm \sqrt{2+\sqrt{8}} = 1+\sqrt{2}\left ( 1 \pm\sqrt{1+\sqrt{2}} \right ) \approx \left\{\begin{matrix} 4.612\\0.217 = \frac{1}{4.612} \end{matrix}\right.\]

The roots are reciprocals, because the left hand side of $(1)$ is symmetric in $x$ and $\frac{1}{x}$, so both roots yield the same RHS, namely $\approx$ 4.612. Therefore, only the larger root is valid.

* yes , by checking only the larger root is valid

 

[Opalg]

Variation on Ifdahl's solution:
[sp]Let $y = x + \frac1x$. Then $\bigl(\sqrt{y+1} + \sqrt y\bigr)\bigl(\sqrt{y+1} - \sqrt y\bigr) = y+1-y = 1$. Therefore $$\sqrt{y+1} - \sqrt y = \frac1{\sqrt{y+1} + \sqrt y} = \frac1x.$$ Now add that equation to the given equation $\sqrt{y+1} + \sqrt y = x$, getting $2\sqrt{y+1} = x + \frac1x = y.$ Square both sides: $4(y+1) = y^2$. That is a quadratic, with solutions $y = 2(1\pm\sqrt2).$ Discard the negative solution (because we need to have $\sqrt y$ available). Then $$x + \frac1x = y = 2(1+\sqrt2).$$ Multiply by $x$, getting another quadratic equation $x^2 - 2(1+\sqrt2)x + 1 = 0$. As Ifdahl correctly points out, only the larger solution of this equation, $x = 1+\sqrt2 + \sqrt{2(1+\sqrt2)} \approx 4.6116$, satisfies the original equation.[/sp]