The Complexity's Simplicity

As it is known and proven, the sum of a quadratic equation's roots can be expressed as $\frac{-b}{a}$ , and their product, as $\frac{c}{a}$ . Hence:

$r + s = \frac{-b}{a}$ and $r \times s = \frac{c}{a}$

So $r = \frac{-b}{a} - s \Rightarrow r = \frac{-b -a.s}{a}$ . As $r > 0$ :

$\frac{-b -a.s}{a} > 0 \Rightarrow -b - a.s > 0 \Rightarrow -b > a.s \Rightarrow b < -a.s$

As $a > 0$ and $s > 0$ , "-a.s" is negative, so "b" is a negative number. Repeating this process with $r \times s$ :

$r = \frac{c}{a.s} \Rightarrow \frac{c}{a.s} > 0$

As $a.s > 0$ (because $a > 0$ ), "c" is a positive number.

Therefore, $b + c < 0$ is true only if $-b > c$ or $\lvert b \rvert > \lvert c \rvert$ . As r and s are decimal numbers, where $0 < r < 1$ and $0 < s < 1$ , their sum is always greater than their product. So:

$r + s > r.s \Rightarrow \frac{-b}{a} > \frac{c}{a} \Rightarrow -b > c\therefore \LARGE \boxed{b + c < 0}$