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Math Help - Just 3 problems I can't figure out guys!

  1. #1
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    Just 3 problems I can't figure out guys!



    Could you please help me solve these 3? The total assignment is 16 questions but these are the 3 troubling me.


    1)

    x^2/3 + 3x^1/3 - 18 = 0

    Solution set is ( , )

    2)

    (1/y-9)^2 + 4(1/y-9) - 21 = 0

    3)

    a= x^2 + 16x -7

    ( , )
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  2. #2
    MHF Contributor
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    Quote Originally Posted by knowledgekick View Post
    1) x^2/3 + 3x^1/3 - 18 = 0
    You can convert this to quadratic "form":

    . . . . . \left(x^{\frac{1}{3}}\right)^2\, +\, 3\left(x^{\frac{1}{3}}\right)\, -\, 18\, =\, 0

    Factor this quadratic, and then solve the two exponential equations which result.

    Quote Originally Posted by knowledgekick View Post
    2) (1/y-9)^2 + 4(1/y-9) - 21 = 0
    This is the same sort of thing, except that the resulting equations will be rational equations.

    . . . . . \left[\left(\frac{1}{y\, -\, 9}\right)\, +\, 7\right]\left[\left(\frac{1}{y\, -\, 9}\right)\, -\, 3\right]\, =\, 0

    Quote Originally Posted by knowledgekick View Post
    3) a= x^2 + 16x -7
    All you can do with this one, since you don't have a value for "a", is to move the unknown over to the right-hand side:

    . . . . . 0\, =\, x^2\, +\, 16x\, -\, 7\, -\, a

    Then apply the Quadratic Formula, with the Formula's a = 1, b = 16, and c = -7 - a. (The two a's are different, obviously.)

    If you get stuck, please reply showing how far you have gotten. Thank you!
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