Algebra

3x^4-31x^2+28=0 Solve the equation??

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  1. you will need to use the quadratic equation

    google search if you don't know

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    posted by Do
  2. let y = x^2
    then
    3 y^2 -31 y + 28 = 0
    (3y-28)((y-1) = 0
    y = 1 or 28/3
    so
    x^2 = 1
    x = 1 or -1
    x^2 = 28/3
    x = 2 sqrt(7/3) or -2sqrt(7/3)

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    posted by Damon
  3. let x^2 = a
    so we have
    3a^2 - 31a + 28 = 0
    (3a-28)(a-1) = 0
    a = 28/3 or a = 1
    so
    x^2 = 28/3
    x = ± 2√7/√3 or a = ±1
    or
    x = ± 2√21/3 , a = ±1

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    posted by Reiny

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