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$${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$

Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.

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  • $\begingroup$ Please show us how you obtained a degree 6 equation. $\endgroup$ – Namaste Jul 26 '18 at 20:28
  • $\begingroup$ @user579048 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Aug 10 '18 at 23:37
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Let's see how insurmountable this sextic equation is.

Start by deriving the equation. First multiply by $\sqrt{x}$, square both sides, and isolate the remaining radical:

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

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

Square again, expand and collect to get our monster:

$x^6-2x^5-x^4+2x^3+x^2=0$

Clearly since $x$ must be positive (certainly not zero, which the fractional radicands in the original equations would forbid), the factor $x^2$ can be divided out. We are down to degree $4$:

$x^4-2x^3-x^2+2x+1=0$

Next observe that this quartic equation has the following property:

(Linear coefficient)/(Cubic coefficient) $=c$

(Constant)/(Quartic coefficient)$=c^2$

When this occurs, the roots of the quartic equation occur in pairs giving the product $c$ and thus we must have a factorization:

$(x^2+ax+c)(x^2+bx+c)$

Here, $c=-1$ and so:

$x^4-2x^3-x^4+2x+1=0=(x^2+ax-1)(x^2+bx-1)$

Expanding the right side and matching like terms leads to two independent equations, thus:

$a+b=-2; b=-2-a$

$ab=1$

Then $a(-2-a)=1, a^2+2a+1=0$ and we have the root $a=-1$. Thus $b=-1$ and our polynomial is now reduced to a single, squared factor:

$(x^2-x-1)^2=0$

And so, from the quadratic formula,

$x=\frac{1+\sqrt{5}}{2}$

Which the reader can verify, checks out! Note that in the checking process we should find $(x^2-1)/x=1$.

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HINT

We can try with

$${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$

$$\sqrt {x+1}{\sqrt {x- 1\over x}} + {\sqrt{x-1\over x}} = x$$

$$(\sqrt {x+1}+1){\sqrt {x- 1\over x}} = x$$

$${\sqrt {x- 1\over x}} = \frac{x}{\sqrt {x+1}+1}\frac{\sqrt {x+1}-1}{\sqrt {x+1}-1}=\sqrt {x+1}-1$$

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

and from here we can square to eliminate the square roots.

Recall to check at the end the conditions for the existence related to the original equation

  • ${{x^2- 1\over x}}\ge 0$

  • ${{x- 1\over x}}\ge 0$

  • $x\neq 0$

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  • $\begingroup$ Can I see the radical eliminated equation? It looks like it would be degree only 4 making it more parsimonious than my (and the OP's) development. $\endgroup$ – Oscar Lanzi Jul 27 '18 at 10:00
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    $\begingroup$ @OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$. $\endgroup$ – gimusi Jul 27 '18 at 10:03
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Checking other answers:

Note that $\frac{x^2-1}{x}\ge 0;\frac{x-1}{x}\ge 0$ and $x>0$ imply $x>1$.

If $x^2=x+1 \ \ (1)$, then: $$ x^2-1=x \Rightarrow \frac{x-1}{x}=\frac{1}{x+1},\\ {\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x \ \ \ \ \ \ \ \ \ \ (2) \ \ \ \ \ \Rightarrow \\ {\sqrt {(x+1)- 1\over x}} + {\sqrt{1\over x+1}} = \sqrt{x+1} \Rightarrow \\ 1+\frac1{\sqrt{x+1}}=\sqrt{x+1} \Rightarrow \\ \sqrt{x+1}+1=x+1 \Rightarrow \\ x+1=x^2.$$ So, $(1)$ and $(2)$ are equivalent.

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  • $\begingroup$ How d you magically get $x^2=x+1$? $\endgroup$ – Oscar Lanzi Jul 27 '18 at 17:32
  • $\begingroup$ @OscarLanzi, the point was to support the given two answers and show the golden mean is indeed the root of the given equation. $\endgroup$ – farruhota Jul 27 '18 at 18:35
  • $\begingroup$ See my introductory phrase. $\endgroup$ – Oscar Lanzi Jul 28 '18 at 1:09
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    $\begingroup$ @OscarLanzi, it is fine, thank you. $\endgroup$ – farruhota Jul 28 '18 at 13:52

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