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Spivak: typo or not?

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data

    In problem 1.18.b of Calculus, he says:

    .

    (Bold mine). I assume that is supposed to say in fact x2+bx+c>0 ?
    However, I am having trouble proving that for all choices of b and c.

    If I complete the square on x2+bx+c>0 I get

    x2+bx+c = (x+b/2)2 + k > 0 => k > 0.

    So I have that k = c - b2/4 > 0. But I don't know that this is true for all b,c.
     
  2. jcsd
  3. Feb 8, 2012 #2

    Dick

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    You were given b^2-4c<0. So 4c-b^2>0. So c-b^2/4>0. It's not true for all b,c. But it is true for all b,c such that b^2-4c<0, isn't it?
     
  4. Feb 8, 2012 #3
    Arg. Yes. I guess so. Bad wording IMO. But I should have seen that. :facepalm:
     
  5. Feb 8, 2012 #4

    Dick

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    "If fact" is bad wording. Probably distracted you enough to miss the point.
     
  6. Feb 8, 2012 #5
    I have a semi-related question(s): We have shown that if b2 - 4c < 0,then x2 + bx + c > 0 for all x.

    Now, if I turn around and say the "reverse": If x2 + bx + c > 0 for all x, then b2 - 4c < 0.

    Question 1: Is what I have just written called the "converse" of the original statement? And I do not think that in general the converse follows .... I would have to prove it. Correct?
     
  7. Feb 8, 2012 #6

    Mark44

    Staff: Mentor

    Yes, that's the converse of the original statement, and it is true that the converse does not necessarily have to be true when the original statement is true.

    Here's a simple example:

    If x = -2, then x2 = 4

    The converse is: If x2 = 4, then x = -2, which is not true.
     
  8. Feb 8, 2012 #7
    Thank you Mark44. Good stuff. And as it turns out, it happens to be true in this case:

    Proof:

    Assume x2 + bx + c > 0 for all x,

    completing square:

    x2 + bx + c = (x + b/2)2 + (c - b2/4) > 0

    and since the term in bold must be ≥ 0 then we have b2 - 4c < 0.

    Thanks Dick and Mark!
     
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