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mskyeg

  • 2 years ago

What are the real or imaginary solutions of the polynomial equation? x3 = 216

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  1. mskyeg
    • 2 years ago
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    Possible answers

  2. hartnn
    • 2 years ago
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    \(x^3-216=0 \\ x^3-6^3=0 \\ \text{now apply }a^3-b^3 formula\)

  3. mskyeg
    • 2 years ago
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    how do i do that

  4. mskyeg
    • 2 years ago
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    like what does it mean

  5. hartnn
    • 2 years ago
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    know the formula ?

  6. mskyeg
    • 2 years ago
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    like what do i plug intohe formula

  7. mskyeg
    • 2 years ago
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    *into the

  8. hartnn
    • 2 years ago
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    a=x,b=6

  9. mskyeg
    • 2 years ago
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    how does that help all i get is x^3 - 216

  10. hartnn
    • 2 years ago
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    thats because u don't know that formula

  11. mskyeg
    • 2 years ago
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    oooh i thought that WAS the formula

  12. hartnn
    • 2 years ago
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    \(\large a^3-b^3=(a-b)(a^2+ab+b^2)\)

  13. hartnn
    • 2 years ago
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    now put a=x,b=6

  14. hartnn
    • 2 years ago
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    and then equate it to 0

  15. hartnn
    • 2 years ago
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    confused ?

  16. hartnn
    • 2 years ago
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    \((x-6)(x^2+6x+36)=0 \\ (x-6)=0,(x^2+6x+36)=0 \\ x=6,\text{use quadratic formula for x^2+6x+36=0}\)

  17. hartnn
    • 2 years ago
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    Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two roots of x are: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

  18. mskyeg
    • 2 years ago
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    I tried but i got a really weird answer

  19. hartnn
    • 2 years ago
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    what u got, a,b,c as ?

  20. mskyeg
    • 2 years ago
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    a=6 b=-3+i square root 54 c=-3-i square root 54

  21. hartnn
    • 2 years ago
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    ???

  22. mskyeg
    • 2 years ago
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    oh a=1 b=6 and c=36

  23. hartnn
    • 2 years ago
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    yes

  24. hartnn
    • 2 years ago
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    so two other values of x are ?

  25. hartnn
    • 2 years ago
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    one obviously is x=6

  26. mskyeg
    • 2 years ago
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    what about the other two

  27. hartnn
    • 2 years ago
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    thats what u have u work out using \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\) u have a,b,c values correct.

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