Algebraically find all the real solutions for the polynomial?

2x^6+28=15x^3

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  • Philip
    Lv 6
    3 weeks ago

    2x^6 +28 = 15x^3...(1).;                                                                                            Put x^3 = u. Then (1) ---> 2u^2+28 = 15u, ie., 2u^2-15u+28= 0, ie.,(2u-7)(u-4) = 0. Then u = (7/2) or 4 and x = (7/2)^(1/3) or 4^(1/3) = 1.518294486 or 1.587401052.;

    Put a = u^(1/3). Then a^3 = [u^(1/3)]^3 = u = x^3 and x^3-a^3 = 0...(2).;

    For u = (7/2), a = (7/2)^(1/3) = 1.518294486.. For u = 4, a = 2^(1/3) = 1.587401052;

    Each a-value occurs in a form (2) eqn equivalent to (x-a)(x^2+ax+a^2) = 0...(3).

    We already got 2 x-roots of eqn (1). 2 x-roots are located in each quadratic eqn (3).

    x^2+ax+a^2 = 0 ---> 2x = -a(+/-)D, where D^2 = a^2 -4a^2 = -3a^2 = (ia*rt3)^2.;

    Then x = (1/2)[-a(+/-)a*i*rt3] = (a/2)[-1(+/-)i*rt3].;

    For a = (7/2)^(1/3), we have x = (1/2)(1.528294486)[-1(+/-)i*rt3].;

    For a.=.......2^(1/3), we have x = (1/2)(1.587401052)[-1(+/-)i*rt3].

    I got a little too excited and decided to solve for all 6 roots of (1). Some of you may

    appreciate my enthusiasm, others may groan. Oh well, that's life.  

  • ?
    Lv 7
    3 weeks ago

    2x⁶ + 28 = 15x³

    2x⁶ - 15x³+ 28 = 0.........[1]

    Let u = x³ then [1] can be rewritten as

    2u² - 15u + 28 = 0

    // Solve for u using the quadratic formula

    .......- (-15) ± √((-15)² - 4(2)(28))

    u = ------------------------------------

    ................2(2)

    .................15 ± √1

    ..........u = -----------

    ......................4

    ..........u = ⁷⁄₂, 4

    // Now, back substitute for u = x³

    x³ = ⁷⁄₂, 4

    x = ∛⁷⁄₂ or  x = ∛4

    x = 1.518, or 1.587.......................ANS

  • 3 weeks ago

    2x^6 + 28 = 15x^3

    2x^6 - 15x^3 + 28 = 0

    (x^3 - 4) (2 x^3 - 7) = 0

    Real solutions:

    x = 2^(2/3)

    x = (7/2)^(1/3)

  • 3 weeks ago

    Rewrite that as:

    2x^6 - 15x^3 + 28 = 0

    Then treat (x^3) as a variable:

    2(x^3)^2 - 15(x^3) + 28 = 0

    Factor by grouping, notiting that 2*28 = 56 = (-7)(-8) and (-7) + (-8) = -15:

    2(x^3)^2 - 7(x^3) - 8(x^3) + 28 = 0

    [2(x^3) - 7](x^3) - (4)[2(x^3) - 7] = 0

    [2(x^3) - 7] [x^3 - 4] = 0

    That's true when either 2x^3 = 7 or x^3 = 4.

    The only real values are cuberoot(7/2) and cuberoot(4).

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  • 2x^6 - 15x^3 + 28 = 0

    x^3 = (15 +/- sqrt(225 - 4 * 2 * 28)) / 4

    x^3 = (15 +/- sqrt(225 - 224)) / 4

    x^3 = (15 +/- sqrt(1)) / 4

    x^3 = (15 +/- 1) / 4

    x^3 = 16/4 , 14/4

    x^3 = 4 , 3.5

    x = 4^(1/3) , 3.5^(1/3)

  • cosmo
    Lv 7
    3 weeks ago

    Let u = x^3                                 

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