Find all real numbers such that $\sqrt{x-\frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$

Question

Find all real numbers such that

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

My attempt to the solution :

I tried to square both sides and tried to remove the root but the equation became of 6th degree.Is there an easier method to solve this?

Answer 1

Let's multiply $$ \sqrt{x-1/x} + \sqrt{1 - 1/x} = x\tag{1} $$ by $(\sqrt{x-1/x} - \sqrt{1 - 1/x})$. Then we get $$ x-1=x(\sqrt{x-1/x} - \sqrt{1 - 1/x}) $$ $$ \sqrt{x-1/x} - \sqrt{1 - 1/x}=1-1/x\tag{2} $$ Sum up $(1)$ and $(2)$ to get $$ 2\sqrt{x-1/x}=x-1/x+1 $$ Now make the substitution $y=\sqrt{x-1/x}$. The rest is clear.

Answer 2

Multiply with $\sqrt{x-1/x}-\sqrt{1-1/x}$ to get $$x-1=(x-\frac1x)-(1-\frac1x) =x\left(\sqrt{x-1/x}-\sqrt{1-1/x}\right)$$ so $$1-\frac1x = \sqrt{x-1/x}-\sqrt{1-1/x}$$ and by adding $$ x+1-\frac1x =2\sqrt{x-\frac1x}$$ Now with $z=x-\frac1x$, this is simply $$z+1=2\sqrt z,$$ a quadratic in $\sqrt z$ with (double) solution $\sqrt z=1$, so $z=1$ and we need to solve $$ x-\frac1x=1$$ i.e. $$ x^2-x-1=0.$$ This has solutions $x=\frac{1\pm\sqrt 5}2$, but only the positive value is possible.

Answer 3

Hints:

$$x=\sqrt{x-\frac1x}+\sqrt{1-\frac1x}\implies x\sqrt x=\sqrt{x^2-1}+\sqrt{x-1}\implies$$

$$ x^3=x^2+x-2+2\sqrt{x^3-x^2-x+1}\implies(x^3-x^2-x+2)^2=4(x^3-x^2-x+1)$$

Well, now you can put $\,w:=x^3-x^2-x+1\;$ and solve

$$(w+1)^2=4w\iff (w-1)^2=0\iff w=1\;\ldots\;\text{and etc.}\ldots$$

Answer 4

Clearly, $x\ne0$

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

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

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

$$\implies \sqrt{x-1}\frac{x+1-x}{\sqrt{x+1}-1}=x\sqrt x\text{ (rationalizing the numerator) }$$

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

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

Squaring we get, $$x+x-1+2\sqrt{x^2-x}=x^2+x\iff 2\sqrt{x^2-x}=x^2-x+1$$

Putting $x^2-x=a^2,a^2=2a-1\implies (a-1)^2=0\implies a=1$