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Math Help - Need to clarify a quadratic equation (simple)

  1. #1
    Senior Member Mukilab's Avatar
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    Need to clarify a quadratic equation (simple)

    Is it true that in this equation:

    \frac{x}{99}=(a-b)(a+b)

    To make the fraction a whole number (x being a multiple of 99) it is not possible to make the (a-b) and (a+b) make a multiple of 99, with a and b being single digit numbers only.

    I don't think so myself. The only possible situation would be used two digit numbers e.g. a as 10 and b as 1. However this would defy the question.
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  2. #2
    MHF Contributor

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    Why would you want a- b or a+ b to be a multiple of 99? The left side is not likely to be!

    The right side is (a- b)(a+ b)= a^2- b^2, a difference of squares.
    Neither 1 nor 2 is a difference of squares but 3 is: you could have x= 3(99)= 297, a= 2, b= 1. 4 is not a difference of squares but 5 is: you could have x= 5(99)= 495, a= 3, b= 2.

    Am I misunderstanding your question?
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  3. #3
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    Quote Originally Posted by Mukilab View Post
    Is it true that in this equation:

    \frac{x}{99}=(a-b)(a+b)

    To make the fraction a whole number (x being a multiple of 99) it is not possible to make the (a-b) and (a+b) make a multiple of 99, with a and b being single digit numbers only.

    I don't think so myself. The only possible situation would be used two digit numbers e.g. a as 10 and b as 1. However this would defy the question.
    1. I assume that x > 0, a > 0, b > 0. Thus a > b.

    2. If a-b=1 then x = 99(a+b)

    3. Therefore you'll get these single digit results:

    (a,b) = (a, a-1),~ a \in  \{2,3,...,9\}

    4. If a-b=3 then x = 3 \cdot 99(a+b)

    5. Therefore you'll get these single digit results:

    (a,b) = (a, a-3),~ a \in  \{4,5,...,9\}

    6. Use the same way to dtermine the values of a and b if

    a-b = 5
    a-b = 7

    7. If a-b=9 then you have to use at least two-digit-numbers.
    Last edited by earboth; March 30th 2011 at 12:44 AM.
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