Solving Spivak's Problem 1.18.b: Showing x2+bx+c>0 for all x

[Saladsamurai]

Homework Statement

In problem 1.18.b of Calculus, he says:

Suppose b²-4c<0. Show that there are no numbers x such that x²+bx+c=0. If fact, x²+bx+c>0 for all x. Hint: Complete the square.

.

(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.

 

[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?

 

[Saladsamurai]

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

 

[Dick]

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

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

 

[Saladsamurai]

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?

 

[Mark44]

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.

 

[Saladsamurai]

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.

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!