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mskyeg Group Title

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

  • one year ago
  • one year ago

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  1. mskyeg Group Title
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    Possible answers

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

    • one year ago
  3. mskyeg Group Title
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    how do i do that

    • one year ago
  4. mskyeg Group Title
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    like what does it mean

    • one year ago
  5. hartnn Group Title
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    know the formula ?

    • one year ago
  6. mskyeg Group Title
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    like what do i plug intohe formula

    • one year ago
  7. mskyeg Group Title
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    *into the

    • one year ago
  8. hartnn Group Title
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    a=x,b=6

    • one year ago
  9. mskyeg Group Title
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    how does that help all i get is x^3 - 216

    • one year ago
  10. hartnn Group Title
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    thats because u don't know that formula

    • one year ago
  11. mskyeg Group Title
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    oooh i thought that WAS the formula

    • one year ago
  12. hartnn Group Title
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    \(\large a^3-b^3=(a-b)(a^2+ab+b^2)\)

    • one year ago
  13. hartnn Group Title
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    now put a=x,b=6

    • one year ago
  14. hartnn Group Title
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    and then equate it to 0

    • one year ago
  15. hartnn Group Title
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    confused ?

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

    • one year ago
  17. hartnn Group Title
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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: \(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

    • one year ago
  18. mskyeg Group Title
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    I tried but i got a really weird answer

    • one year ago
  19. hartnn Group Title
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    what u got, a,b,c as ?

    • one year ago
  20. mskyeg Group Title
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    a=6 b=-3+i square root 54 c=-3-i square root 54

    • one year ago
  21. hartnn Group Title
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    ???

    • one year ago
  22. mskyeg Group Title
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    oh a=1 b=6 and c=36

    • one year ago
  23. hartnn Group Title
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    yes

    • one year ago
  24. hartnn Group Title
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    so two other values of x are ?

    • one year ago
  25. hartnn Group Title
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    one obviously is x=6

    • one year ago
  26. mskyeg Group Title
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    what about the other two

    • one year ago
  27. hartnn Group Title
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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.

    • one year ago
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