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

What are the real roots?

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    In cases like this, and *after* answers have been posted, it may be a better idea to ask a new question rather than radically changing the old one. – 2017-02-03

3 Answers 3

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[ EDIT ] Original hint remains in the edit history, the following applies to the re-edited question.

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

Taking the inverses on both sides (since $x\ne0\,$):

$$ \require{cancel} \begin{align} \frac{1}{x} & = \cfrac{1}{\sqrt{x-\frac{1}{x}} + \sqrt{1-\frac{1}{x}}} \;\cdot\; \cfrac{\sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}}}{\sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}}} \\[3pt] & = \cfrac{\sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}}}{\left(x-\cancel{\frac{1}{x}}\right) - \left(1-\cancel{\frac{1}{x}}\right)} = \cfrac{\sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}}}{x-1} \end{align} $$

$$ \iff\quad\;\; \cfrac{x-1}{x} = \sqrt{x-\frac{1}{x}} - \sqrt{1-\frac{1}{x}} \tag{2} $$

Subtracting $(1)-(2)$ gives:

$$ x - \frac{x-1}{x} = 2 \sqrt{1-\frac{1}{x}} \;\;\iff\;\; x - 1 + \frac{1}{x} = 2 \sqrt{1 - \frac{1}{x}} \tag{3} $$

Multiplying by $x\,$, squaring, and noting that the equation becomes of the form $(a+1)^2=4a\,$:

$$ (x^2-x+1)^2=4(x^2-x) \;\;\iff\;\; (x^2 - x - 1)^2 = 0 \tag{4} $$

Since the RHS of $(1)$ is positive, the only qualifying root $x \gt 0$ of $(4)$ is $\;x = \cfrac{1 + \sqrt{5}}{2}\,$.

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$\sqrt {g(x)}\ge 0$. Always! This is the way the square root operation is defined.

If $\sqrt {x - \frac 1x} + \sqrt {1 - \frac 1x} =0$ then $x - \frac 1x = 0$ and $1 - \frac 1x = 0$

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    Is there a reason why the equation under the radical equals the sum? In other words, why does $x-\frac 1x$ equal zero? Did you take the square root of both sides? – 2017-02-02
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    No, I think you missed my point. $\sqrt {x - \frac 1x} \ge 0$ and $\sqrt {1 - \frac 1x} \ge 0$ by the definition of square roots. If $\sqrt {x - \frac 1x} + \sqrt {1 - \frac 1x} =0$ then both $\sqrt {x - \frac 1x}$ and $\sqrt {1 - \frac 1x} =0$ because if either is greater than 0, then the sum cannot equal 0. Finally, if $\sqrt {1 - \frac 1x} =0$ then $1 - \frac 1x =0$ – 2017-02-02
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    Ah okay I understand, thanks! – 2017-02-02
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Leaving aside questions of choice of square roots, the only way $x - 1/x$ can be equal to $1 - 1/x$ is $x=1$.