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anonymous

  • one year ago

Determine how many, what type, and find the roots for f(x) = x4 + 21x2 − 100.

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  1. Hayleymeyer
    • one year ago
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    lets take t=x^2

  2. welshfella
    • one year ago
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    its degree 4 so how many roots will there be?

  3. anonymous
    • one year ago
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    4

  4. Hayleymeyer
    • one year ago
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    so we have the equation t^2 +21t-100 =0

  5. welshfella
    • one year ago
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    yea

  6. Hayleymeyer
    • one year ago
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    Looking at the discriminant we have discriminant = \[\frac{ -21\pm \sqrt{21^2-4(1)(-100)} }{ 2}\]

  7. welshfella
    • one year ago
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    - which can be factored

  8. welshfella
    • one year ago
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    no need this can be factored

  9. Hayleymeyer
    • one year ago
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    simplifying the discriminant we get - t=4 or t= -25

  10. Hayleymeyer
    • one year ago
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    we know that x^2 =t so \[x=\sqrt{t}\]

  11. anonymous
    • one year ago
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    ok

  12. Hayleymeyer
    • one year ago
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    we have 2 values of t :) one is positive and one is negative :) what do u think can we put negative value of t in this function to get x- \[x=\sqrt{t}\]?

  13. anonymous
    • one year ago
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    idk to be honest

  14. welshfella
    • one year ago
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    t = 4 gives 2 values for x 2 and - 2

  15. Hayleymeyer
    • one year ago
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    ok well can the square of any number be negative ?

  16. welshfella
    • one year ago
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    now the square roots of -25 are imaginary do you know what they are?

  17. welshfella
    • one year ago
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    yes - you introduce the operator i which stands for the square root of -1.

  18. anonymous
    • one year ago
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    5i?

  19. welshfella
    • one year ago
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    so sqrt -15 = -5i and 5 i

  20. welshfella
    • one year ago
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    *sqrt -25

  21. welshfella
    • one year ago
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    and theres your 4 roots 2, -2 , 5i and -5i

  22. Hayleymeyer
    • one year ago
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    well \[\sqrt{-25} \] does not exists cause -25 is negative so we r left with t=4 puttin t=4 in the equation \[x=\sqrt{t}\]we get x=2 and x=-2 :)

  23. welshfella
    • one year ago
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    probably you haven't come to complex and imaginary numbers yet. They are not real numbers but they do exist in math.

  24. anonymous
    • one year ago
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    i have i know what they are...somewhat

  25. welshfella
    • one year ago
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    Yes - mathematicians introduced them because some problems could not be solved using real numbers alone.

  26. welshfella
    • one year ago
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    Hayleymeyer obviously hasn't been taught them yet.

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