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Math Help - quadratic equation problem no.3

  1. #1
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    quadratic equation problem no.3

    quadratic equation problem no.3
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  2. #2
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    You have been given k is an >positive< integer right??
    So use it

    Notice that a and b are rationals and roots of x^2 + 3x + k = 0.

    And if k>=3 then 9 - 4k < 0 and hence \sqrt {9-4k} does not exist. This means You have to check only for k=1 and k=2, Notice that k=1 \Rightarrow \sqrt {9-4k} = \sqrt 5 which is irrational. But this cant hold because we have been told that roots are rationals. Thus now check for k=1........ And then claim it is the only solution.

    All done
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  3. #3
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    that means
    9-4k > 0
    k > 0
    therefore 0<k<9/4
    k=2 or 1 {rejected as x = sqrt5 [irrational] when k = 1}
    thanks!
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