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Thu, 25 Apr 2019
Sometimes it matters how you get there
Katara was given the homework exercise of rationalizing the denominator of $$\frac1{\sqrt2+\sqrt3+\sqrt5}$$ which she found troublesome. You evidently need to start by multiplying the numerator and denominator by !!-\sqrt2 + \sqrt 3 + \sqrt 5!!, obtaining $$ \frac1{(\sqrt2+\sqrt3+\sqrt5)}\cdot \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{-\sqrt2 + \sqrt 3 + \sqrt 5} = \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{(-2 +3 + 5 + 2\sqrt{15})} = \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{6 + 2\sqrt{15}} $$ and then you go from there, multiplying the top and bottom by !!6 - 2\sqrt{15}!!. It is a mess. But when I did it, it was much quicker. Instead of using !!-\sqrt2 + \sqrt 3 + \sqrt 5!!, I went with !!\sqrt2 + \sqrt 3 - \sqrt 5!!, not for any reason, but just at random. This got me: $$ \frac1{\sqrt2+\sqrt3+\sqrt5}\cdot \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{\sqrt2 + \sqrt 3 - \sqrt 5} = \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{(2 +3 - 5 + 2\sqrt{6})} = \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{2\sqrt{6}} $$ with the !!2+3-5!! vanishing in the denominator. Then the next step is quite easy; just get rid of the !!\sqrt6!!: $$ \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{2\sqrt{6}}\cdot \frac{\sqrt6}{\sqrt6} = \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12} $$ which is correct. I wish I could take credit for this, but it was pure dumb luck. [Other articles in category /math] permanent link |