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Math Help - Find all real solutions- Please help

  1. #1
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    Post Find all real solutions- Please help

    I have a problem that I am stuck on. I have been asked to solve the problem by finding all real solutions. The problem is: x^6-9x^3+8=0

    I thought that I should use the ax^2+bx+c=0 but the 6th power is throwing me off.

    If anyone could help, I would really appreciate it!

    Thanks,
    Ashley
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  2. #2
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    Hi

    You are right
    x^6-9x^3+8=0

    Let y = x^3 then y^2 = (x^3)^2 = x^6

    And the equation becomes y^2-9y+8=0

    Solve for y and then find x
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  3. #3
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    Hi! Thanks for the help!

    The only question I have is- After doing the quadratic formula, is it possible to have to answers for "y"? And if I do have to answers (I have y=3 and y=-4) how do I plug that in for "x"?

    Thanks again! Sorry if that is confusing!
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  4. #4
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    Quote Originally Posted by aperkin2 View Post
    I have y=3 and y=-4
    Are you sure about your "solutions" for y ?

    You must solve y^2-9y+8=0
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  5. #5
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    Oh, probably not. After doing the quadratic formula I ended up with--

    y= -1 + and - the square root of 49 all over 2
    So I thought that would be -8/2 and 6/2
    That would be -4 and 3.

    I have messed up somewhere I am sure. Thanks again!
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  6. #6
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    The solutions are given by the formula

    y = \frac{-b \pm \sqrt{\Delta}}{2a}

    Therefore y = \frac{9 \pm \sqrt{49}}{2}
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  7. #7
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    Oh I see where I messed up! I was plugging in the wrong value for b. Thank you so much for your help! It has been great!
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