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Math Help - interesting quadratic equation problem

  1. #1
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    Smile interesting quadratic equation problem

    hi everyone.I'm just wondering ,if you have the roots of a quadratic equation is it possible to determine it's coefficients,i suspect i can determine b and c while a will have a range of possible values.What do you think?
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    If you have the equation
    x^2 + 2x - 3 = 0 You can factor the quadratic to get
    (x - 1)(x + 3) = 0 and hence the roots
    x = 1 and x = -3
    Conversely, if you have the roots x = 1 and x = -3
    you can recover the quadratic equation by forming (x - 1)(x + 3) = 0.

    The same is true if the roots are complex,

    for example suppose that the roots are

    5 + 3i and 5 - 3i then the quadratic equation is

    (x - 5 - 3i)(x - 5 + 3i) = 0
    which simplifies to

    x^2 - 10x + 34 = 0

    Cheers,
    Adarsh
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  3. #3
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    thanks for replying

    thanks for replying.Using the method your using suggests a is one always.I'm trying to find a way of calculating at least one of the values of a that is not one.a is the first coefficient
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    So lets take it together

    1)GIVEN: Roots are given

    2) ALREADY FOUND : you found single b and c

    3) NEXT FINDING: You found a range of "a" for a constant value of roots and other coefficients

    If you agree to all the above steps than I dont agree with you
    Now give a second thought and
    tell me incase still you dont agree with it
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  5. #5
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    Quote Originally Posted by dynamo View Post
    hi everyone.I'm just wondering ,if you have the roots of a quadratic equation is it possible to determine it's coefficients,i suspect i can determine b and c while a will have a range of possible values.What do you think?
    You can find a monic polynomial with the given roots P_1(x)=(x-r_1)(x-r_2), then all polynomials with these roots are of the form:

     <br />
P(x)=\alpha P_1(x)<br />

    CB
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  6. #6
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    thanks for replying

    thanks for replying
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