How should I solve the equation $\sqrt{x-\frac 1x}+\sqrt{1-\frac 1x}=x$

Question

I could square both sides of the equation, but that ends up giving me a cubic to solve. What I need is a beginning approach to solve such questions, not the whole answer.

Thanks

Answer 1

Note that $x = 0$ is clearly not a solution. Observe that: \begin{align*} &\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x \\ &\Rightarrow \left(\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\right)\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\ &\Rightarrow \left(\left(x - \frac{1}{x}\right) - \left(1 - \frac{1}{x}\right)\right) = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\ &\Rightarrow x - 1 = x\left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) \\ &\Rightarrow \sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} = 1 - \frac{1}{x} \end{align*}

Adding both together yields: \begin{align*} &\left(\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}}\right) + \left(\sqrt{x - \frac{1}{x}} - \sqrt{1 - \frac{1}{x}}\right) = x + \left(1 - \frac{1}{x}\right) \\ &\Rightarrow 2\sqrt{x - \frac{1}{x}} = \left(x - \frac{1}{x}\right) + 1 \end{align*} We substitute $y = \sqrt{x - \frac{1}{x}}$, and we see that: \begin{align*} 2y = y^2 + 1 \Rightarrow (y - 1)^2 = 0 \Rightarrow y = 1 \end{align*} It remains to solve $\sqrt{x - \frac{1}{x}} = 1$.

Answer 2

Hint: After squaring two times we get $$- \left( {x}^{2}-x-1 \right) ^{2}=0$$

Answer 3

Square $\sqrt{x-1/x}=x-\sqrt{1-1/x}$ to get $x^2-x+1=2x\sqrt{1-1/x}$. Square again,

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

which yields $x=\frac{1\pm\sqrt5}{2}$, of which only the positive root is the true solution as required by the original equation

$$x=\frac{1+\sqrt5}{2}$$