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

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.

  • $\begingroup$ Please show us how you obtained a degree 6 equation. $\endgroup$ – amWhy 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/… $\endgroup$ – user Aug 10 '18 at 23:37

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:



Square again, expand and collect to get our monster:


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 $4$:


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:


Here, $c=-1$ and so:


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

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


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:


And so, from the quadratic formula,


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

| cite | improve this answer | |


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

  • ${{x^2- 1\over x}}\ge 0$

  • ${{x- 1\over x}}\ge 0$

  • $x\neq 0$

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ @OscarLanzi squaring 2 times we obtain $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0$. $\endgroup$ – user Jul 27 '18 at 10:03

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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ @OscarLanzi, it is fine, thank you. $\endgroup$ – farruhota Jul 28 '18 at 13:52

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.