Results 1 to 4 of 4

Thread: Quadratic expression

  1. #1
    Newbie
    Joined
    May 2011
    Posts
    14

    Quadratic expression

    Find the condition that

    $\displaystyle ax^2 + 2hxy + by^2$

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


    Not a clue on this one....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    $\displaystyle \displaystyle \begin{align*}(y - mx)(my + x) &= my^2 + xy - m^2xy - mx^2 \\ &= -mx^2 + (1 - m^2)xy + my^2\end{align*}$

    If this is equal to $\displaystyle \displaystyle ax^2 + 2hxy + by^2$ then $\displaystyle \displaystyle a = -m, 2h = 1 - m^2, b = m$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2011
    Posts
    14
    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 $\displaystyle b = m$ and $\displaystyle a = - m$ , $\displaystyle a + b = 0$." and $\displaystyle 2h = 1 - m^2$ . The answer given at the back of the book is $\displaystyle a + b = 0$ . But the $\displaystyle 2h = 1 - m^2$ can't simply be ignored , can it ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2011
    Posts
    14
    I've got it !

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

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

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

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

    Then since $\displaystyle m$ satisfies both the equations

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

    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

    $\displaystyle \frac{a}{b } = \frac{-2h}{2h} = \frac{b}{a} $
    $\displaystyle \Rightarrow \frac{a}{b } = -1 = \frac{b}{a}$
    $\displaystyle \Rightarrow a + b = 0$ Q.E.D
    Last edited by mathlover14; May 31st 2011 at 09:35 AM. Reason: Got the full solution
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic expression
    Posted in the Algebra Forum
    Replies: 8
    Last Post: Jun 1st 2011, 06:49 PM
  2. divisibility of quadratic expression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 16th 2011, 03:21 AM
  3. Quadratic Expression?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Apr 16th 2010, 02:22 AM
  4. [SOLVED] Quadratic expression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Apr 7th 2010, 10:32 AM
  5. quadratic expression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Apr 6th 2010, 05:29 PM

Search Tags


/mathhelpforum @mathhelpforum