# Homework Help: Spivak: typo or not?

1. Feb 8, 2012

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. Feb 8, 2012

### Dick

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?

3. Feb 8, 2012

Arg. Yes. I guess so. Bad wording IMO. But I should have seen that. :facepalm:

4. Feb 8, 2012

### Dick

"If fact" is bad wording. Probably distracted you enough to miss the point.

5. Feb 8, 2012

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?

6. Feb 8, 2012

### 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.

7. Feb 8, 2012

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!