Japan | A Nice Square Root Algebra Problem | Math Olympiad

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[@عبدالواسع-س8م]

Great job !

[@brendaandalistairhunter9593]

√108+√36=√3.√36+√36= √36(√3+1)
Similarly the denominator=√36(√3-1)
Then multiply √3+1/√3-1 by √3+1/√3+1 etc

[@toveirenestrand3547]

sqrt108 = sqrt(2*2*3*3*3) = 6sqrt3.
(sqrt108 + sqrt6)/(sqrt108 - sqrt36) = (6sqrt3 + 6)/(6sqrt3 - 6) = [6(sqrt3 + 1)]/[6(sqrt3 - 1)] = (sqrt3 + 1)/(sqrt3 - 1) = [(sqrt3 + 1)/(sqrt3 - 1)]*[(sqrt3 + 1)/(sqrt3 + 1)]
= (sqrt3 + 1)^2/[(sqrt3)^2 - 1] = (3 + 2sqrt3 +1)/2 = (4 + 2sqrt3)/2 = 2(2 + sqrt3)/2 = 2 + sqrt3.