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

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In summary, -The converse of the original statement is that if x2 + bx+c > 0 for all x, then b2 - 4c < 0.-The converse of the original statement is that if x2 + bx+c > 0 for all x, then b2 - 4c < 0.
  • #1
Saladsamurai
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Homework Statement



In problem 1.18.b of Calculus, he says:

Spivak said:
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.
 
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  • #2
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
Arg. Yes. I guess so. Bad wording IMO. But I should have seen that. :facepalm:
 
  • #4
Saladsamurai said:
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.
 
  • #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?
 
  • #6
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
Mark44 said:
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!
 

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

How do I approach solving Spivak's Problem 1.18.b?

To solve this problem, you will need to use basic algebraic principles such as factoring and the quadratic formula. It is important to carefully read and understand the question before attempting to solve it. Begin by identifying the given values of b and c and then use them to find the discriminant, which will help determine the nature of the quadratic equation.

What is the discriminant and how is it used in this problem?

The discriminant is a term used in the quadratic formula and is represented by the expression (b^2 - 4ac). In this problem, the discriminant is used to determine the nature of the quadratic equation. If the discriminant is positive, the quadratic equation will have two distinct real roots and the expression x2+bx+c will always be greater than 0 for all values of x. If the discriminant is zero, the quadratic equation will have one repeated real root and the expression x2+bx+c will always be greater than or equal to 0. If the discriminant is negative, the quadratic equation will have two complex roots and the expression x2+bx+c will not always be greater than 0.

Can I use any other methods besides the quadratic formula to solve this problem?

Yes, you can also use factoring to solve this problem. If the quadratic equation can be factored into two linear factors, then it will always be greater than 0 for all values of x. However, using the quadratic formula is a more reliable method as it can be applied to any quadratic equation, regardless of whether it can be factored or not.

What is the significance of showing that x2+bx+c>0 for all values of x?

This means that the quadratic equation will always have a positive y-intercept and will never cross or intersect the x-axis. In other words, the graph of the quadratic equation will always lie above the x-axis and will never have any real roots, making it a "happy" parabola. This is important in many mathematical applications, such as in optimization problems.

What are some tips for successfully solving Spivak's Problem 1.18.b?

Some tips include carefully reading and understanding the question, identifying the given values and their relationships, using the quadratic formula or factoring method accurately, and double-checking your final answer by plugging in different values for x to ensure that the expression x2+bx+c is always greater than 0. It is also important to show all steps and clearly explain your reasoning to demonstrate a thorough understanding of the problem.

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