Luxembourg - Math Olympiad Question You should know this trick 

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sqrt9 = 3, sqrt8 = 2sqrt2 3 - 2sqrt(2). want to make it some squared quantity to get rid of the secondary sqrt. lets say x = sqrt2, x^2 = 2. we create both of these terms, and are left with + 1 for the constant. x^2 - 2x + 1, which is a perfect square: = (x-1)^2 --> sqrt(2) - 1 Let's try to expand this further. Want (A + B*sqrt(K))^2 = M + Usqrt(K); Where M, U, and K are given. = (A^2 + KB^2) + 2ABsqrt(K) Match coefficients: A^2 + KB^2 = M 2AB = U --> AB = U/2 --> B = [U/2]/A = U/2A (Isolate for B) Now plug B back into eq 1: A^2 + K* (U/2A)^2 - M = 0 A^4 - MA^2 + KU^2 / 4 = 0 [Multiply out by A^2] A^2 = [M +/- sqrt(M^2 - KU^2)]/2 [So if M^2 - KU^2 < 0, we only have imaginary answers, which doesn't seem to work here] Our best bet is taking the '+' version given that it is a squared value, and must be positive: A^2 = [M + sqrt(M^2 - KU^2)]/2 A = sqrt[(M + sqrt(M^2 - KU^2))/2] (Let's just take the positive version here for ease of use also) B = U/2A ---------------------------------------------------------------------------- We test this out with our starting problem: -- A = sqrt[(M + sqrt(M^2 - KU^2))/2] -- B = U/2A -- A + B*sqrt(K) = sqrt(M + Usqrt(K)), Ie, our result of A + Bsqrt(K) should be equal to our initial result, with sqrt [3 - 2sqrt(2)] = sqrt(2) - 1 M = 3, U = -2, K = 2 A = sqrt[(3 + sqrt(9 - 2*4))/2] = sqrt[(3 + 1)/2] = sqrt(4/2) = sqrt(2) B = -2 / 2sqrt(2) = -1/sqrt2 A + B*sqrt(K) = sqrt2 - 1