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Math Help - quadratic formula

  1. #1
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    quadratic formula

    The Quadratic Formula Explained says:

    So, while factoring may not always be successful, the Quadratic Formula can always find the solution.


    I'm trying to use the Quadratic Formula to solve
    1.26x^2-186.56x+34041.94=0 but my 4ac is much bigger than my b^2 so everything under the radical evaulates to a negative number. How can I solve this?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by erinspice View Post
    The Quadratic Formula Explained says:



    I'm trying to use the Quadratic Formula to solve
    1.26x^2-186.56x+34041.94=0 but my 4ac is much bigger than my b^2 so everything under the radical evaulates to a negative number. How can I solve this?
    it IS possible for a quadratic to have no real solutions. Look at x^2 + 1 = 0. This is never true for any real number x. Have you learned about complex numbers, and complex solutions? If so, the quadratic formula can still find those
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    it IS possible for a quadratic to have no real solutions. Look at x^2 + 1 = 0. This is never true for any real number x. Have you learned about complex numbers, and complex solutions? If so, the quadratic formula can still find those
    Yeah, but I need a real solution. My robot won't know what to do with a distance of xi (see this thread)! I must have made an error somewhere, because I know there has to be a real solution to my problem.
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by erinspice View Post
    The Quadratic Formula Explained says:



    I'm trying to use the Quadratic Formula to solve
    1.26x^2-186.56x+34041.94=0 but my 4ac is much bigger than my b^2 so everything under the radical evaulates to a negative number. How can I solve this?
    If you wish to check your answers, even those which have complex roots, download the MHF QuickSolver. The link to it is in my signature.
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  5. #5
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    Complex zeros

    If one sketches the quadratic function, one can see that graph of the function does not at any point intersect x-axis thus there is no real zero. However, for polynom of degree 2 there must neccesarily be two zeros. Because they are not real, they must be complex, which I guess in your case is not neccesary to calculate.
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