Solve: $\sqrt{x-\frac1x}-\sqrt{1-\frac1x}=1-\frac1x$ for $ x\neq0$

Question

$$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}{=1-\frac{1}{x}}, x\neq0$$ So my idea is that I moved the second sqrt to the RHS and left the first on the LHS. I then squared everything and ended up with $\frac{x^2-1}{x}=2-\frac{3}{x}+\frac{1}{x^2}$. I don't know what to do next? Can somebody give a hint?

Answer 1

Multiply everything by $x^2$: $$x^3-x=2x^2-3x+1$$ Rearrange: $$x^3-2x^2+2x-1=0.$$ Note that $x-1$ is a factor: $$(x-1)(x^2-x+1)=0$$ Now hopefully you can solve it.

Answer 2

squaring two times and foctorizing we get $$-\frac{(x-1)^2 \left(x^2-x-1\right)^2}{x^4}=0$$ if we do so we get $$x-\frac{1}{x}=(1-\frac{1}{x})^2+1-\frac{1}{x}+2(1-\frac{1}{x})\sqrt{1-\frac{1}{x}}$$ the square root is missing !! and you must square again!!!!!!!

Answer 3

Note that $x=1$ is a solution. Else, let $A = \sqrt{x-\frac{1}{x}}$ and $B = \sqrt{1-\frac{1}{x}}$

then $A-B = \frac{x-1}{x}$ and $A^2-B^2 = x-1$, then dividing we get $A+B = x$

Adding we get $2\sqrt{x-\frac{1}{x}} = 1+x - \frac{1}{x}$ or $2A = 1+A^2 \Rightarrow A=1$ or $x^2-x+1 = 0$

Answer 4

The domain gives $-1\leq x<0$ or $x\geq1$.

1. $-1\leq x<0$.

We obtain: $$\sqrt{\frac{(1-x)(x+1)}{-x}}=\frac{\left(\sqrt{1-x}\right)^2}{x}+\sqrt{\frac{1-x}{-x}}$$ or $$\sqrt{\frac{x+1}{-x}}=\frac{\sqrt{1-x}}{x}+\frac{1}{\sqrt{-x}}$$ or $$-\sqrt{-x(x+1)}=\sqrt{1-x}-\sqrt{-x}$$ or $$\sqrt{-x}=\sqrt{1-x}+\sqrt{-x(1+x)}$$ or $$-x=1-x-x-x^2+2\sqrt{x(x^2-1)}$$ or $$x^2+x-1=2\sqrt{x(x^2-1)},$$ which is impossible because $x^2+x-1<0$ for $-1\leq x<0.$

2. $x\geq1$.

We obtain: $$\sqrt{\frac{(x-1)(x+1)}{x}}=\frac{x-1}{x}+\sqrt{\frac{x-1}{x}},$$ which gives $$x=1$$ or $$\sqrt{\frac{x+1}{x}}=\frac{\sqrt{x-1}}{x}+\sqrt{\frac{1}{x}},$$ which is $$\sqrt{(x+1)x}=\sqrt{x-1}+\sqrt{x}$$ or $$x^2+x=2x-1+2\sqrt{(x-1)x}$$ or $$x^2-x-2\sqrt{(x-1)x}+1=0$$ or $$\left(\sqrt{x^2-x}-1\right)^2=0$$ or $$x^2-x-1=0,$$ which gives $$x=\frac{1+\sqrt{5}}{2}$$ and we got the answer: $$\left\{1,\frac{1+\sqrt5}{2}\right\}$$