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Math Help - 2 questions on my math homework

  1. #1
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    2 questions on my math homework

    I can't seem to figure these two questions out. A little assistance please?

    1)

    x^2/3 + 2x^1/3 - 63 = 0

    The solution set is ( , )

    2)

    For each given pair of numbers find a quadratic equation with integral coefficients that has the numbers as its solutions.

    -4i, 4i
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by knowledgekick View Post
    I can't seem to figure these two questions out. A little assistance please?

    1)

    x^2/3 + 2x^1/3 - 63 = 0

    The solution set is ( , )

    2)

    For each given pair of numbers find a quadratic equation with integral coefficients that has the numbers as its solutions.

    -4i, 4i
    Is your first question  x^{\frac{2}{3}} + 2x^{\frac{1}{3}} - 63 = 0???
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  3. #3
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    Quote Originally Posted by harish21 View Post
    Is your first question  x^{\frac{2}{3}} + 2x^{\frac{1}{3}} - 63 = 0???
    Yes, thank you.
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by knowledgekick View Post
    Yes, thank you.
    Substitute u = x^{1/3} on the LHS:

     {u^2} + 2u - 63 = 0

     u^2+ 2u = 63

    Add 1 to both sides:

    u^2+2u+1 = 64

    (u+1)^2 = 64

    Take the square root of both sides:

     u+1 = \pm 8

     u = -9 OR  u = 7

    Substitute back for u = x^{1/3}

    x^{\frac{1}{3}}= -9 OR  x^{\frac{1}{3}} = 7

    Cube both sides:

    x = -729 OR  x = 343

    Now try plugging these values into your question and check!

    Was that helpful?
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  5. #5
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    Yes! Thank you very much! I'm clueless on what to do for the other problem though. Any idea?
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by knowledgekick View Post
    Yes! Thank you very much! I'm clueless on what to do for the other problem though. Any idea?
    Your solutions are given as 4i and -4i

    So, you can set up the equation as:

    (x-4i)(x+4i)=0

    Remember, for imaginary number i,

    i^{2} = -1
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  7. #7
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    I got x^2 + 16 for that. I think it's right. Thanks a lot!

    I have a couple more that are bothering me now. If you could help that would be great. If not, I understand...

    1)

    x^2/3 + 3x^1/3 - 28 = 0

    Solution set is ( , )

    I tried doing this the same way you did the first one but I'm not getting it!

    2)

    Find the vertex

    a= x^2 + 18x - 8
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  8. #8
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by knowledgekick View Post
    I got x^2 + 16 for that. I think it's right. Thanks a lot!

    I have a couple more that are bothering me now. If you could help that would be great. If not, I understand...

    1)

    x^2/3 + 3x^1/3 - 28 = 0

    Solution set is ( , )

    I tried doing this the same way you did the first one but I'm not getting it!

    2)

    Find the vertex

    a= x^2 + 18x - 8
    x^{2/3} + 3x^{1/3} - 28 = 0

    Substitute u = x^{1/3} on the LHS:

    You have

    u^{2} + 3u -28 = 0

    u^{2} +7u -4u -28 = 0

    u(u+7)-4(u+7) = 0

    (u+7)(u-4) = 0

    therefore u = -7 OR u = 4

    substitute u = x^{1/3} and solve as done before
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  9. #9
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by knowledgekick View Post
    I got x^2 + 16 for that. I think it's right. Thanks a lot!

    I have a couple more that are bothering me now. If you could help that would be great. If not, I understand...

    1)

    x^2/3 + 3x^1/3 - 28 = 0

    Solution set is ( , )

    I tried doing this the same way you did the first one but I'm not getting it!

    2)

    Find the vertex

    a= x^2 + 18x - 8
    (2)

    a= x^2 + 18x - 8 is a parabola..........(1)

    The equation of a parabola can be written as:

    y = ax^{2}+bx+c.........(2)

    And the vertex of the parabola is given by :vertex = \frac{-b}{2a}

    Compare equation (1) and equation (2), your values are:

    b = 18 and a = 1

    So your vertex is......
    Last edited by harish21; April 8th 2010 at 11:22 AM.
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  11. #11
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by knowledgekick View Post
    -9
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