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Math Help - Need Little Help

  1. #1
    kap
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    Need Little Help

    Hello Everybody,

    I need help in solving this question. The equation is:

    5x^2 + 4x + 1 \equiv 0

    The question is to:

    determine all the solutions in Z(17) and in Z(20).
    [In Z(17): 9 and 14, in Z(20): none ]

    NOTE::
    The Z(17) is suppose to be the whole numbers symbol with subscript 17 but i don't know how to do it in Latex. The same applies to Z(20).

    Thanks for your help.
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  2. #2
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    Quote Originally Posted by kap View Post
    Hello Everybody,

    I need help in solving this question. The equation is:

    5x^2 + 4x + 1 \equiv 0

    The question is to:

    determine all the solutions in Z(17) and in Z(20).
    [In Z(17): 9 and 14, in Z(20): none ]

    NOTE::
    The Z(17) is suppose to be the whole numbers symbol with subscript 17 but i don't know how to do it in Latex. The same applies to Z(20).

    Thanks for your help.
    5x^2 + 4x + 1 \not \equiv 0
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  3. #3
    kap
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    Could you please explain your answer to me a little bit.
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  4. #4
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    Quote Originally Posted by kap View Post
    Could you please explain your answer to me a little bit.
    You have no real solution but complex.
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  5. #5
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    Hello, kap!

    I never studied modulo arithmetic formally,
    . . but I just "invented" a method . . .


    Solve: . 5x^2 + 4x + 1 \:\equiv\: 0 \text{ (mod 17)}

    We have: . 5x^2 + 4x + 1 \:\equiv\:0\text{ (mod 17)}


    Multiply by 7: . 35x^2 + 28x +  7 \:\equiv\:0\text{ (mod 17)}

    . . . . . . . . . . . x^2 + 11x + 24 \:\equiv\:0\text{ (mod 17)}

    . . . . . . . . . . . (x + 8)(x+3) \:\equiv\:0\text{ (mod 17)}


    Therefore: . \begin{Bmatrix}\;x &\equiv& -8 &  \equiv & 9 & \text{(mod 17)}\;\\<br />
x &\equiv& -3 & \equiv & 14 & \text{(mod 17)}\end{Bmatrix}

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