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

I tried sq both sides by taking 1 in RHS but it didn't worked out well...

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    both Terms are the same? – 2017-02-26

4 Answers 4

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Squaring both sides:

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

Squaring again we get:

$x^{4} - 2x^{3} - x^{2} + \frac{4}{x^{2}} + 6x -\frac{4}{x} - 3 = 4(\frac{1}{x^{2}} + x - \frac{1}{x} - 1)$

We can reduce this to:

$x^{4} -2x^{3} -x^{2} +2x + 1 = 0$

Notice that this factorises into $(x^{2} - x -1)^{2} = 0$

$\Rightarrow x = \frac{1 \pm \sqrt{5}}{2}$

Note this is the 'golden ratio' $\phi = \frac{1 + \sqrt{5}}{2}$ with the property that: $\frac{1}{\phi} = \phi - 1$

Since we have squared our initial equation thus introducing extra solutions, we must test each of these in the original equation.

This leads us to reject $x = \frac{1 - \sqrt{5}}{2}$

Therefore $x= {\frac{1 +\sqrt{5}}{2}}$

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    I don't see which one you're referring to – 2017-02-26
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    Ok thnks for the help.. – 2017-02-26
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    Ah I had a typo with $x^{2} = \frac{1 \pm \sqrt{5}}{2}$ instead of $x = \frac{1 \pm \sqrt{5}}{2}$ – 2017-02-26
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HINT: after squaring one times we get $$2\sqrt{x-\frac{1}{x}}\sqrt{1-\frac{1}{x}}=x^2-x+\frac{2}{x}-1$$ can you finish this? squaring this one more times we get $$4\left(x-\frac{1}{x}\right)\left(1-\frac{1}{x}\right)=\left(x^2-x+\frac{2}{x}-1\right)^2$$ expanding the left Hand side we obtain $$4\,x-4-4\,{x}^{-1}+4\,{x}^{-2}$$ and the right Hand side is given by $${x}^{4}-2\,{x}^{3}+6\,x-{x}^{2}-3+4\,{x}^{-2}-4\,{x}^{-1}$$ have you got this? Bringung all together we obtain this equation $$0=x^4-2x^3-x^2+2x+1$$

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    Shouldn't it be +$\frac{2}{x}$ on the RHS? – 2017-02-26
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    yes you have right sorry corrected , i have worked today the whole time – 2017-02-26
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    I did this before but was not able to complete it.. – 2017-02-26
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    i will do the rest – 2017-02-26
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    It is becoming very complicated after squareing 2nd time – 2017-02-26
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    yes i have posted this for you – 2017-02-26
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    Got it.. but there is a slight error in ur last eqn.. istesd of -10x it should be +2x.. – 2017-02-26
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    yes i have corrected it i was the whole day by the Matholympiad in Leipzig – 2017-02-26
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Divide by $x$: $$\sqrt{\frac1x-\frac1{x^3}}+\sqrt{\frac1{x^2}-\frac1{x^3}}=1$$ Change $t=1/x$: $$\sqrt{t-t^3}+\sqrt{t^2-t^3}=1$$ Square and rearrange: $$2\sqrt{t^3-t^4-t^5+t^6}=t+t^2-2t^3-1$$ Square and rearrange again: $$4t^6-4t^5-4t^4+4t^3=4t^6-4t^5-3t^4+6t^3-t^2-2t+1$$ Finally we get $$t^4+2t^3-t^2-2t+1=0$$ That is, $$(t^2+t-1)^2=0$$

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$x\geq1$ and $1$ is not root.

Hence we can rewrite our equation in the following form: $$x-1=x\left(\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}\right)$$ or $$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}=1-\frac{1}{x}$$ and with the given we obtain $$2\sqrt{x-\frac{1}{x}}=x+1-\frac{1}{x}$$ or $$x-\frac{1}{x}-2\sqrt{x-\frac{1}{x}}+1=0$$ or $$\left(\sqrt{x-\frac{1}{x}}-1\right)^2=0$$ or $$x^2-x-1=0,$$ which gives $x=\frac{1+\sqrt5}{2}$.