Algebra 2

"The Discriminant; Equations in Quadratic Form"

Solve over the complex numbers.

1/e + 1/(square root of e) - 6 = 0

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asked by Mary
  1. I would let √e = x , then
    1/x^2 + 1/x - 6 = 0
    multiply by x^2
    1 + x - 6x^2 = 0
    6x^2 - x - 1 = 0
    (3x+1)(2x-1) = 0
    x = -1/3 or x = 1/2
    √e = -1/3 or √e = 1/2
    e = 1/9 or e = 1/4

    since we "squared" we have to verify all answers.
    if e = 1/9
    LS = 1/(1/9) + 1/√(1/9) - 6
    = 9 + 3 - 6 = 6
    RS = 0, so e = 1/9 does not work

    if e = 1/4
    LS = 4 + 2 - 6 = 0 = RS

    so e = 1/4

    BTW, most texts and authors would avoid using e as a variable, since e is generally accepted as the Euler constant

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    posted by Reiny
  2. Thank you so much! I've been stuck on this problem for over 2 hours

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    posted by Mary

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