Can you Solve Oxford University Pure Mathematics Entrance Exam ?

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  • Опубліковано 30 жов 2024

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  • @key_board_x
    @key_board_x 37 хвилин тому

    √{3√[3.(x + 1).√{3.(x + 1)}]} = 9 → where: (x + 1) ≥ 0 → x ≥ - 1
    3√[3.(x + 1).√{3.(x + 1)}] = 81
    √[3.(x + 1).√{3.(x + 1)}] = 27
    3.(x + 1).√{3.(x + 1)} = 729
    (x + 1).√{3.(x + 1)} = 243
    √{3.(x + 1)} = 243/(x + 1)
    3.(x + 1) = 243²/(x + 1)²
    3 = 243²/(x + 1)³
    3.(x + 1)³ = 243²
    (x + 1)³ = 19683
    (x + 1)³ - 19683 = 0
    (x + 1)³ - 27³ = 0 → recall: a³ - b³ = (a - b).(a² + ab + b²)
    [(x + 1) - 27].[(x + 1)² + 27.(x + 1) + 27²] = 0
    (x + 1 - 27).(x² + 2x + 1 + 27x + 27 + 729) = 0
    (x - 26).(x² + 29x + 757) = 0
    First case: (x - 26) = 0
    x - 26 = 0
    x = 26
    Second case: (x² + 29x + 757) = 0
    x² + 29x + 757 = 0
    Δ = 29² - (4 * 757) = 841 - 3028 = - 2187 ← negative value, we stop here