Luxembourg - Math Olympiad Question You should know this trick

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  • čas přidán 27. 08. 2024
  • Math Olympiad Question
    Maths Olympiads are held all around the world to recognise students who excel in maths.
    The test is offered at many grade levels and provides them with numerous possibilities to win certifications, awards, and even scholarships for higher studies.
    #math #maths #mathstricks #mathematicsanalysis #algebra #algebratricks #trigonometry #Exponent #simplify #Simplification #Olympiad #OlympiadAlgebra #howtosolvethisproblem #howtosolve #algebraproblems #Exponentproblem #findx #findallvaluesof #powerof #simplemaths
    Luxembourg - Math Olympiad Question You should know this trick

Komentáře • 4

  • @javidmirab-jo6hi
    @javidmirab-jo6hi Před měsícem

    Excellent

  • @darcash1738
    @darcash1738 Před měsícem +1

    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

  • @user-wm5od9xj5i
    @user-wm5od9xj5i Před měsícem

    Great channel

  • @user-id1nm3wc2e
    @user-id1nm3wc2e Před měsícem

    All right