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\(\[\cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \dotsb}}}}\]\)

Guest Jun 3, 2017

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 #1
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Let x = the thing that you typed <-- too lazy XD

\(x =\cfrac{1}{1+\dfrac{1}{2+x}}\\ x + \dfrac{x}{2+x}=1\\\ x(x + 2) + x = x + 2\\ x^2 + 3x - x - 2 = 0\\ x^2 + 2x - 2 = 0\\ x = \dfrac{-(2)\pm\sqrt{2^2 - 4(1)(-2)}}{2(1)}\\ x =\dfrac{-2 \pm 2\sqrt{3}}{2}=-1\pm\sqrt{3}\\ x = \sqrt3 - 1\text{ or }x = -1-\sqrt3\text{(rej.) <-- Do you even think it's a negative number?}\\ x = \sqrt3 - 1\)

MaxWong  Jun 4, 2017
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 #1
avatar+6810 
+1
Best Answer

Let x = the thing that you typed <-- too lazy XD

\(x =\cfrac{1}{1+\dfrac{1}{2+x}}\\ x + \dfrac{x}{2+x}=1\\\ x(x + 2) + x = x + 2\\ x^2 + 3x - x - 2 = 0\\ x^2 + 2x - 2 = 0\\ x = \dfrac{-(2)\pm\sqrt{2^2 - 4(1)(-2)}}{2(1)}\\ x =\dfrac{-2 \pm 2\sqrt{3}}{2}=-1\pm\sqrt{3}\\ x = \sqrt3 - 1\text{ or }x = -1-\sqrt3\text{(rej.) <-- Do you even think it's a negative number?}\\ x = \sqrt3 - 1\)

MaxWong  Jun 4, 2017

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