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Equation of exponents

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Source: Canada 1998

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#1 h Mar 4, 2006, 12:45 PM • 1 Y

Y by Adventure10

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

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#2 h Mar 4, 2006, 2:14 PM • 1 Y

Y by Adventure10

It give
$x(\sqrt{x-\frac 1x}-\sqrt{1-\frac 1x})=x-1$ . Therefore
$\sqrt{x-\frac 1x}=\frac{x+1-\frac 1x }{2}$ Let $y=x-\frac 1x$ , We have $\sqrt y =\frac{y+1}{2}$ . It give y=1.
If find really roots it give $x=\frac{1+\sqrt 5 }{2}$ .

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#3 h Mar 4, 2006, 2:22 PM • 3 Y

Y by nicky-glass, Adventure10, Mango247

$\begin{eqnarray*}x=\sqrt{1-{1\over x}}+\sqrt{x-{1\over x}} &\iff& x-\sqrt{1-{1\over x}}=\sqrt{x-{1\over x}}\\ &\stackrel{\textrm{attention!}}{\implies}& x^2-2x\sqrt{1-{1\over x}}+1-{1\over x}=x-{1\over x}\\ &\iff& x^2-2\sqrt{x^2-x}+1=x\\ &\iff& x^2-x-2\sqrt{x^2-x}+1=0\\ &\iff& (\sqrt{x^2-x}-1)^2=0\\ &\iff& x^2-x=1\\ &\iff& x^2-x-1=0\\ &\iff& x_{1,2}=\frac{1\pm\sqrt{5}}{2}\end{eqnarray*}$

Since by the initial equation $x$ must be positive, the only solution is $x=\frac{1+\sqrt{5}}{2}$

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#4 h Apr 24, 2014, 12:24 AM • 1 Y

Y by Adventure10

Albanian TST 2014

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#5 h Apr 17, 2025, 5:37 PM

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wrote this a while ago but never posted fsr

We claim the answer is $\boxed{x=\dfrac{1+\sqrt{5}}{2}}.$ We proceed as follows.

Since $x=\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}},$ $x-\sqrt{x-\dfrac{1}{x}}=\sqrt{1-\dfrac{1}{x}},$ so $1-\dfrac{1}{x}=\left(\sqrt{1-\dfrac{1}{x}}\right)^2=\left(x-\sqrt{x-\dfrac{1}{x}}\right)^2=x^2+x-\dfrac{1}{x}-2x\sqrt{x-\dfrac{1}{x}},$ so $2x\sqrt{x-\dfrac{1}{x}}=x^2+x-1.$ Clearly, $x\neq 0,$ so $2\sqrt{x-\dfrac{1}{x}}=x+1-\dfrac{1}{x}=\left(\sqrt{x-\dfrac{1}{x}}\right)^2+1,$ so $\left(\sqrt{x-\dfrac{1}{x}}-1\right)^2=\left(\sqrt{x-\dfrac{1}{x}}\right)^2-2\sqrt{x-\dfrac{1}{x}}+1=0,$ and $\sqrt{x-\dfrac{1}{x}}=1,$ so $x-\dfrac{1}{x}=1.$ We now have $x^2-x-1=0,$ so $x=\dfrac{1\pm\sqrt{5}}{2}.$

If $x=\dfrac{1-\sqrt{5}}{2},$ then $x<0<\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}},$ a contradiction. On the other hand, if $x=\dfrac{1+\sqrt{5}}{2},$ then $x=\dfrac{1+\sqrt{5}}{2}=1+\dfrac{\sqrt{5}-1}{2}=\sqrt{\dfrac{1+\sqrt{5}}{2}-\dfrac{\sqrt{5}-1}{2}}+\sqrt{\dfrac{3-\sqrt{5}}{2}}=\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\left(\dfrac{\sqrt{5}-1}{2}\right)}=\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}.$ As such, the only solution is $x=\dfrac{1+\sqrt{5}}{2},$ as desired. $\Box$

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