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Algebra

posted by on .

Can someone tell me how they got the answer?

Find a polynomial equation with real coefficients that has the given roots.
4i, sqrt5
my answer: x^4-22x^2+80=0
correct answer: x^4+11x^2-80=0

  • Algebra - ,

    If 4i is a root, there must have been a -4i
    and if √5 was a root there must have been a -√5

    so the factors were (x-4i)(x+4i)(x+√5)x-√5)
    = (x^2 + 16)(x^2-5)
    = x^4 - 5x^2 + 16x^2 - 80
    = x^4 + 11x^2 - 80

  • Algebra - ,

    (x+4i)(x-4i)(x-sqrt5)(x+sqrt5)
    (x^2 + 16)(x^2-5)
    x^4 +11x^2 - 80


    I don't know what you did.

  • Algebra - ,

    What is that saying?
    something about "great minds ...." .lol

    notice even the posing time was the same

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