Solving A Nice Cubic Equation

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  • Опубліковано 11 гру 2024

КОМЕНТАРІ • 8

  • @LazarAndric-se4cq
    @LazarAndric-se4cq День тому +1

    7:30 you missed the sign of 1/27

  • @robertibatullin8329
    @robertibatullin8329 16 годин тому

    Simplification of x in the 2nd method is a problem in its own right.

  • @stvp68
    @stvp68 16 годин тому

    My motto is no pain, no pain 😅

  • @raghvendrasingh1289
    @raghvendrasingh1289 День тому

    👍
    8 x^3+8 x - 5 = 0
    there is only one sign change in f(x) and no sign change in f(-x) hence by Descartes' rule of signs equation has only one real root and it is positive.
    by RRT we obtain root , x = 1/2

  • @nasrullahhusnan2289
    @nasrullahhusnan2289 23 години тому

    x³+x=⅝ --> x(x²+1)=⅛(4+1)
    =½×¼(4+1)
    =½(1+¼ )
    =½[1+(½)²]
    Thus x=½ 0:13

  • @Don-Ensley
    @Don-Ensley 11 годин тому

    problem
    x³ + x = 5/8
    Cubic formula
    x = a + b
    a³ + b³ = 5/8
    -3ab = 1
    ab = -1/3
    a³ b ³ = -1/27
    b³ = -1/(27 a³)
    a³ -1/(27 a³) = 5/8
    27 (a³)²-(5/8)27 (a³)-1 = 0
    a³ = { (5/8)27 ± √[(5/8)² 27² +4(27)] }/54
    Δ = (5/8)² 27² +4(27)
    = (27)(25•27/64 +256/64)
    = 9•3/(64) (25•27+256)
    = 9•3/(64) 931
    a³ = { (5/8)27 ± (3/8)√[3•931] }/54
    = { 135 ± 21√57} / 432
    = { 45 ± 7√57} / 144
    x = a + b
    = ∛ [ (45 + 7√57) / 144 ] + ∛ [ (45 - 7√57) / 144 ]
    = 1/2
    This is one real root.
    x -1/2 a factor.
    x³ + x - 5/8 = 0
    (x -1/2) x² +(x-1/2) x/2+(5/4)(x-1/2) = 0
    (x-1/2)(x² +x/2 +5/4) = 0
    ZPP and Quadratic
    x = (-1/2 ± √[(1/4)-5]}/2
    = -1/4 ± 1/4 i √19
    = (-1 ± i √19 )/4
    answer
    x ∈ { 1/2,
    (-1 - i √19 )/4,
    (-1 + i √19 )/4 }

  • @VittorioBalbi1962
    @VittorioBalbi1962 День тому

    😂