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Math Help - two quadratic equation problems

  1. #1
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    two quadratic equation problems

    two quadratic equation problems
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by afeasfaerw23231233 View Post
    two quadratic equation problems
    \left\{ {\begin{array}{ll} xy&= 50 \\ x^{\log y}&= 5 \end{array}} \right.
    xy = 50 \implies x = \frac{50}{y} \implies \log x = \log 5 + 1 - \log y

    thus,  x^{\log y} = 5 \implies \log y \log x = \log 5

    therefore \log y (\log 5 + 1 - \log y) = (\log 5 + 1) \log y - (\log y)^2 = \log 5

    \implies (\log y)^2 - (\log 5 + 1)\log y + \log 5 = 0

    setting  u = \log y , then you are just solving a quadratic equation in 1 variable..
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  3. #3
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    thanks but how to do question 2 [ii]? how to show k = p-1
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  4. #4
    Lord of certain Rings
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    My solution is a little involved
    I call \alpha as 'a' and \beta as 'b' to avoid TeX.

    First of all ,if a and b are integers, so are a-1 and b-1.
    By part {i}, they satisfy x^2+(k+2)x+ p = 0'

    So if a-1 and b-1 satisfy x^2+(k+2)x+ p = 0
    then (a-1)(b-1) = p.

    Since p is prime,the only possible solutions are
    (a,b) = (2,p+1),(p+1,2),(0,-p-1),(-p-1,0)

    We first claim that the first two solutions are invalid.

    If either of a or b were 2, then they satisfy x^2 + kx + (p-k-1) = 0
    that means 4+2k+p-k-1= 0 \Rightarrow 3+p+k = 0
    But this is not possible since p>0 and k>0 .

    So look at the remaining solutions,

    Now that means either of a or b is 0.
    Substitute in x^2 + kx + (p-k-1) = 0
    And get k = p-1

    All done again
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