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Find all values of a that satsify the equation \frac{a}{3} + 1 = \frac{a + 3}{a} + 1.

 May 9, 2024
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\(\displaystyle \frac{a}{3} + 1 = \frac{a + 3}{a} + 1\\ \dfrac a3 + 1 = 2 + \dfrac 3a\\ \dfrac a3 - \dfrac3a - 1 = 0\\ \left(\dfrac a3\right)^2 - \dfrac a3 - 1 = 0\\ \dfrac a3 = \dfrac{1 \pm \sqrt 5}2\\ a = \dfrac{3(1 \pm \sqrt 5)}2\)

 

So the values of a are 3 times the golden ratio, or 3 times the negative reciprocal of the golden ratio. Nice!

 May 9, 2024

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