Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

math exercise?

Solve: x^3 - 6x + 8 = 0

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  • Ian H
    Lv 7
    4 weeks ago

    For x^3 – 3px – 2q = 0 with q^2 > p^3 there is one real root 

    x^3 - 6x + 8 = 0 has p = 2, q = -4, so q^2 > p^3 

    Substitution x = s + p/s and multiplying by s^3 yields the quadratic of cubes 

    (s^3)^2 – 2q(s^3) + p^3 = 0 

    s^3 = q ±√(q^2 – p^3) = -4 ±√(8) ~ -1.1715728752538 

    Selecting the real cube root, s ~ -1.0542002205047  

    x = s + 2/s ~ -2.9513730355915..., which we may call r 

    That is the real root; but every cubic has 3 roots. 

     

    The remaining quadratic must be of the form x^2 + rx - 8/r 

    With solutions x = [-r ±√(r^2 +32/r)]/2 

    In this example √[-(r^2 + 32/r)]/2 = + 0.730035681602 

    So the complex conjugate roots are  

    x = 1.4756865177958 ± 0.730035681602 i 

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  • 4 weeks ago

    Could you perhaps have meant for that to be a quadratic rather than a cubic?

    In the off-chance, you have a typo, I'll show you how to factor the quadratic:

    x² - 6x + 8 = 0

    (x - 2)(x - 4) = 0

    Using the zero product rule, if ab=0, then a=0 or b=0:

    x - 2 = 0

    x = 2

    or

    x - 4 = 0

    x = 4

    Answer:

    x = 2 or x = 4

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  • alex
    Lv 7
    4 weeks ago

    x=-2.9514       

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