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Math Help - Quadratic expression

  1. #1
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    Quadratic expression

    Find the condition that

    ax^2 + 2hxy + by^2

    is resolvable into linear factors of the form (y - mx) and (my + x).


    Not a clue on this one....
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  2. #2
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    \displaystyle \begin{align*}(y - mx)(my + x) &= my^2 + xy - m^2xy - mx^2 \\ &= -mx^2 + (1 - m^2)xy + my^2\end{align*}

    If this is equal to \displaystyle ax^2 + 2hxy + by^2 then \displaystyle a = -m, 2h = 1 - m^2, b = m.
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  3. #3
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    But the question says that there must be be only ONE single condition , not three . But from the three given by you...it can be converted into two...which are.."Since b = m and a = - m , a + b = 0." and 2h = 1 - m^2 . The answer given at the back of the book is a + b = 0 . But the 2h = 1 - m^2 can't simply be ignored , can it ?
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  4. #4
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    I've got it !

    If  (y - mx) be a factor of  ax^2 + 2hxy + by^2 then it must vanish for  y = mx.

    Putting  y = mx in  ax^2 + 2hxy + by^2 , we get  ax^2 + 2hmx^2 + b(m^2)(x^2) = 0 , or  a + 2hm + bm^2 = 0 .

    Similarly if ( my + x) be a factor of  ax^2 + 2hxy + by^2 then it must vanish for  x = - my.

    Putting  x = - my in  ax^2 + 2hxy + by^2 , we get a(m^2)(y^2) - 2hmy^2 + by^2 = 0 or am^2 - 2hm + b = 0 .

    Then since m satisfies both the equations

    bm^2 + 2hm + a = 0..................... (1)
    and, am^2 - 2hm + b = 0....................... (2)

    The condition which is required for both the above equations to have both the roots common is actually the required condition .

    On taking some determinants we easily get that, the required condition is

                       \frac{a}{b } = \frac{-2h}{2h} = \frac{b}{a}
    \Rightarrow  \frac{a}{b } = -1 = \frac{b}{a}
    \Rightarrow   a + b = 0 Q.E.D
    Last edited by mathlover14; May 31st 2011 at 09:35 AM. Reason: Got the full solution
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