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Math Help - Quadratic equation check

  1. #1
    Newbie
    Joined
    Sep 2007
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    Quadratic equation check

    I can't seem to figure out how to check this problem.

    8x^3 + 1 =0

    I factored it and got

    (2x+1)(4x^2 - 2x + 1)

    So x=-1/2 and x= (1張 sqrt 3)/4

    The first solution -1/2 seems to satisfy the equation, but I can't figure out how to check for the second solution.

    8[(1張 sqrt 3)/4]^3 +1=0

    I multiplied the first two numbers and got
    (-2張 sqrt 3)/2 + 1

    and then I got stuck...

    So, could someone please tell me where to go from here?

    Any help would be greatly appreciated.

    Thanks
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    1,631
    For the second solution

    Actually there are 3 solutions:
    The "second" solution composed of two:
    [1 +i*sqrt(3)]/4
    and [1 -i*sqrt(3)]/4

    It's complicated to check with the plus-or-minus, so let's check with the plus,
    x = [1 +i*sqrt(3)]/4

    8x^3 +1 = 0
    8{[1 +sqrt(-3)]/4}^3 +1 =? 0

    We can use the formula (a+b)^3 = a^3 +3(a^2)(b) +3(a)(b^2) +b^3
    but that is complicated too, or very long.
    So we use (a+b)^3 = (a+b)(a+b)^2

    {[1 +sqtrt(-3)]/4}^2
    = (1/16){1 +2sqrt(-3) -3}
    = (1/16){-2 +2sqrt(-3)}
    = (1/8){-1 +sqrt(-3)}

    Multiply that by [1 +sqrt(-3)]/4,
    = (1/8){-1 +sqrt(-3)} * [1 +sqrt(-3)]/4
    = (1/32){-1*1 -1*sqrt(-3) +sqrt(-3) -3}
    = (1/32){-4}
    = -1/8

    So,
    8x^3 +1 = 0
    8{[1 +sqrt(-3)]/4}^3 +1 =? 0
    8{-1/8} +1 =? 0
    -1 +1 =? 0
    0 =? 0
    Yes, so, OK, [1 +i*sqrt(3)]/4 is a root.

    Try the x = [1 -i*sqrt(3)]/4.
    It should check too.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by mi986 View Post
    I can't seem to figure out how to check this problem.

    8x^3 + 1 =0

    I factored it and got

    (2x+1)(4x^2 - 2x + 1)

    So x=-1/2 and x= (1張 sqrt 3)/4

    The first solution -1/2 seems to satisfy the equation, but I can't figure out how to check for the second solution.

    8[(1張 sqrt 3)/4]^3 +1=0

    I multiplied the first two numbers and got
    (-2張 sqrt 3)/2 + 1

    and then I got stuck...

    So, could someone please tell me where to go from here?

    Any help would be greatly appreciated.

    Thanks
    <br />
(1+ i \sqrt{3})^3 = 1 + 3(\pm i \sqrt{3})+ 3(\pm i \sqrt{3})^2+(\pm i \sqrt{3})^3<br />

    .............. <br />
= 1 \pm 3 i \sqrt{3} - 9 \mp 3 i \sqrt{3} = -8<br />

    RonL
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  4. #4
    Newbie
    Joined
    Sep 2007
    Posts
    17
    Ah, thank you both so much. I figured out where I went wrong! Instead of <br />
\mp<br />
    I used both times when I cubed <br />
1 \pm  i \sqrt{3}
    Last edited by mi986; September 23rd 2007 at 05:04 PM. Reason: edited for terrible grammar...
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