Small clarification Spivak Chapter 1, 18(b)

Question

Suppose that ${b}^{2} - 4c < 0$. Show that there are no numbers $x$ satisfying ${x}^{2} + bx + c = 0$; in fact, ${x}^{2} + bx + c > 0$ for all $x$. Hint; Complete the square.

There is one answer here.

Solution (per Spivak)

We have
${x}^{2} + bx + c = \left (x + \frac{b}{2} \right)^{2} + \left(c - \frac{b^{2}}{4} \right) \ge c - \frac{b^{2}}{4} $

Analysis
$\left (x + \frac{b}{2} \right)^{2} $ is $\ge 0$ as any real number $a$, $a^{2} \ge 0$
$\left(c - \frac{b^{2}}{4} \right)$ See question below.

Spivak now states

but $c - \frac{b^{2}}{4} > 0$, so ${x}^{2} + bx + c > 0$ for all $x$

Question

Ignoring the conclusion (included for completeness), what allows him to assert $c - \frac{b^{2}}{4} > 0$?

score 5 · Accepted Answer · answered Oct 25 '21 at 03:52

5

Since the assumption (given condition) was, $$b^2-4c<0\implies c-\frac{b^2}{4}>0.$$

answered Oct 25 '21 at 03:52

Sathvik

3,609

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Thank you so much. I was making it too complicated. – user1115542 Oct 25 '21 at 04:03
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You're welcome :) – Sathvik Oct 25 '21 at 04:03
I think the second line of my ‘analysis’ is incorrect, and I should probably correct that to ‘$>0$ by negated assumption’ – user1115542 Oct 25 '21 at 04:39
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@user1115542 Yes, you are right. – Sathvik Oct 25 '21 at 04:43
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I modified the question to still maintain the spirit of the question, accepting your reply as the accepted answer. – user1115542 Oct 25 '21 at 05:04
Sathvik... would you care to comment on the construct that Spivak uses ie having $c - \frac{b^{2}}{4}$ on either side of the inequality sign. Completing the square accounts for the $c - \frac{b^{2}}{4}$ to the left of the inequality. Thanks. – user1115542 Oct 25 '21 at 20:50
He is using the fact that the square of a real number is non-negative. Hence, $$(x+b/2)^2+c-\frac{b^2}{4}\ge 0+c-\frac{b^2}{4}$$ – Sathvik Oct 26 '21 at 03:12
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As I thought about it, I assumed by doing this, you actually isolate $\left (x + \frac{b}{2} \right)^{2}$. That would more directly meet the proof. And, as you answered in the question, we have proved $c - \frac{b^{2}}{4} > 0$ which "cancels" out on either side of the inequality. If that makes sense :-) – user1115542 Oct 26 '21 at 05:16
Let us continue this discussion in chat. – user1115542 Oct 26 '21 at 06:32

Small clarification Spivak Chapter 1, 18(b)

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