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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.

I asked for clarification a few months ago, here and there is an answer offered here

But, I want to ask specifically about Spivak's style in answering, as it comes up further on again.

His answer is:

We have ${x}^{2} + bx + c = \left( x + \frac{b}{2}\right)^{2} + \left(c - \frac{b^{2}}{4}\right) \ge \left(c - \frac{b^{2}}{4}\right)$ but $\left(c - \frac{b^{2}}{4}\right) > 0 $, so ${x}^{2} + bx + c > 0 $ for all $x$.

It is clearly evident that equality holds for the first two expressions. I would like to understand the use of the expression $\left(c - \frac{b^{2}}{4}\right)$ on either side of the inequality. Hope that makes sense.

user1115542
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1 Answers1

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Here's what he means:

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

Then, because the square of a real number is never negative, we must have

$$\left(x + \frac{b}{2}\right)^2 \ge 0.$$

Hence,

$$\left(x + \frac{b}{2}\right)^2 + \left(c - \frac{b^2}{4}\right) \ge 0 + \left(c - \frac{b^2}{4}\right).$$

Then, since $c - b^2/4 > 0$, it follows that $x^2 + bx + c \ge c - b^2/4 > 0$.

heropup
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