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Homework Help: Random problem from vector cal

  1. Oct 12, 2004 #1
    Can anyone point me in the right direction for this problem? I can't seem to approach it from the right angle:

    Let A, B, C be real numbers such that [tex]A>0[/tex], [tex]B>0[/tex], [tex]AC-B^2>0[/tex].

    a. Prove that a number [tex]Q>0[/tex] exists such that [tex]Ax^2+2Bxy+Cy^2 \geq Q(x^2+y^2)[/tex]

    b. Find the largest possible Q.

    I don't need answers (or rather, I do, but, you know what I mean; my main concern is figuring out how to approach this without my head exploding. Thanks in advance.
  2. jcsd
  3. Oct 12, 2004 #2


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    Are you familiar with the quadratic formula?
  4. Oct 12, 2004 #3
    Solving for x? Solving for y? Solving for Q? I don't understand how that solves anything. Not being a smart-alec; I just don't understand.
  5. Oct 13, 2004 #4


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    I'm just trying to give you hints that you might be able to use to find an answer.

    The LHS of the inequality looks a lot like:

    Similarly, the quadratic formula:
    [tex]\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
    which looks a lot like
  6. Oct 13, 2004 #5


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    Not necessarily the best way but here's how I would do it:
    You want
    [tex]Ax^2+2Bxy+Cy^2 \geq Q(x^2+y^2)[/tex]
    which is the same as
    [tex](A-Q)x^2+ 2Bxy+ (C-Q)y^2\geq 0[/tex].
    The will be true as long as the "discriminant" is not positive:
    [tex]4B^2- 4(A-Q)(C-Q)\leq 0[/tex]
    [tex]4B^2- 4AC+ 4AQ+4CQ- 4Q^2\leq 0[/tex]
    That, of course, is the same as
    [tex]Q^2+ (A+C)Q+ AC- B^2\geq 0[/tex]

    You can use the quadratic formula to find values of Q for which that is true.
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