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Math Help - Determinants Problem

  1. #1
    Newbie Zakaria007's Avatar
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    Post Determinants Problem

    Hey guys,

    I got into quite a predicament lately when I tried to solve this problem. We need to find the value for x such that the determinants below are equal. I tried to expand then equate but got a strange equation, that I couldn't solve. I tried row operations but that failed too. Can you guys please help? Thanks
    Yes, the one on the left is 2*2 and on the right is 3*3.
    Thanks again.
    Attached Thumbnails Attached Thumbnails Determinants Problem-dd.bmp  
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  2. #2
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    I don't understand how the equation you got was strange, you should have included some working so someone could spot where it might have goine wrong.

    Expanding both sides you should end up with

    x(1-x) + 3 = x(x-5) -18 +3x  + 18

    Which is just a quadratic in x. Bring everything onto one side and use the quadratic formula.

    Hope this helps.

    pomp.
    Last edited by pomp; July 9th 2009 at 04:04 AM. Reason: tpyo
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  3. #3
    Newbie Zakaria007's Avatar
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    Lightbulb Quadratic equation solution

    Hey pomp,

    Thanks for the quick response, my question lyes in the quadratic equation itself. When I solved it I got
    2x^2 - 3x - 3 = 0
    and when I tried to solve it by factorization, I couldn't end up with any solution.
    The same thing happened with the quadratic equation (ending up with a negative sign under the root). I'll really appreciate the help. Thanks again.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Zakaria007 View Post
    Hey pomp,

    Thanks for the quick response, my question lyes in the quadratic equation itself. When I solved it I got
    2x^2 - 3x - 3 = 0
    and when I tried to solve it by factorization, I couldn't end up with any solution.
    The same thing happened with the quadratic equation (ending up with a negative sign under the root). I'll really appreciate the help. Thanks again.
    f(x)=2x^2 - 3x - 3 has two real roots so you made a mistake by using the formula x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}. Precisely b^2-4ac=33 here, which is not a negative number.
    Good luck.
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  5. #5
    Newbie Zakaria007's Avatar
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    Smile Final result

    OK man, I get my mistake. I RUSHED!
    The real solution is 9. Thanks guys.
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  6. #6
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    Sorry but that is incorrect. Arbolis pointed out that the equation has 2 real roots.

    f(x) = 2x^2 - 3x -3 = 0

    Then by the quadratic formula:

    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    Here we have a = 2 b = -3 c = -3 so

    x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 2 \times (-3)}}{2 \times 2}

    x = \frac{3 \pm \sqrt{9 + 24}}{4}

    x = \frac{3 + \sqrt{33}}{4}

    and

    x = \frac{3 - \sqrt{33}}{4}

    Neither of these give 9 as an answer. My advice is to always plug your answer back in and see if it satisfies your original equation.

    If you plug 9 into the 2x2 matrix the resulting determinant is -69, whereas the determinant of the 3x3 is 64.
    Also, if you simply plug 9 into your quadratic equation you see that you get 132 and not 0.
    Last edited by pomp; July 10th 2009 at 06:06 AM. Reason: calculation error
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  7. #7
    Newbie Zakaria007's Avatar
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    Post the value of x

    OK I RUSHED AGAIN! Thanks pomp.
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