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Math Help - Quadratic Equation

  1. #1
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    Quadratic Equation

    Solve y


    3*y^(1/2)+8*y^(-1/2)-10=0

    the answer is 16/9 and 4.

    what i did:

    root y = t
    y= t^2

    3t^2+8/t^2-10=0
    3t^4-10t^2+8=0
    (3t^2-4)(t^2-2)=0

    t=16/9 or root 2

    which is not what i want.
    i want t^2=16/9 and 4

    thanks for helping!
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  2. #2
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    Quote Originally Posted by BabyMilo View Post
    3*y^(1/2)+8*y^(-1/2)-10=0
    Multiply through by sqrt[y] to get:

    . . . . . 3y\, +\, 8\, -\, 10\sqrt{y}\, =\, 0

    This is a quadratic in \sqrt{y}:

    . . . . . 3\left(\sqrt{y}\right)^2\, -\, 10\sqrt{y}\, +\, 8\, =\, 0

    Factor this as you would factor 3x^2 - 10x + 8:

    . . . . . (3\sqrt{y}\, -\, 4)(\sqrt{y}\, -\, 2)\, =\, 0

    Then:

    . . . . . \sqrt{y}\, =\, \frac{4}{3}\, \mbox{ or }\, \sqrt{y}\, =\, 2

    Can you take it from there?
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  3. #3
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    3*y^(1/2)+8*y^(-1/2)-10=0
    put y^(1/2) = t

    3t + 8/t -10 = 0
    3t^2 -10t +8 = 0
    (3t - 4)(t-2) = 0

    t = 4/3 or 2
    t^2 = 16/9 or 4
    y = 16/9 or 4
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  4. #4
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    thanks both of you but does anyone know why my method has gone wrong?
    thanks
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  5. #5
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    You set sqrt[y] = t, but substituted t^2 for sqrt[y].
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  6. #6
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    lol i see where i went wrong now!
    thanks!
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