Inequalities: NYJC 2024 Promo Q2

(a)

Method 1:

$$x^2+4x+9=(x+2)^2+5$$ Since $(x+2)^2\geq 0$, $(x+2)^2+5>0$.

Therefore, $x^2+4x+9$ is always positive for all real values of $x$.

Method 2:

$$b^2-4ac=4^2-4(1)(9)=-20$$ Since the discriminant $<0$ and that the coefficient of $x^2$ is positive, therefore, $x^2+4x+9$ is always positive for all real values of $x$.

(b)

$$\begin{aligned} \frac{(x^2+4x+9)(x+2)^2}{x^2-6x+7}&\leq 0\\ \text{Since }x^2+4x+9&>0\\ \frac{(x+2)^2}{x^2-6x+7}&\leq 0\\ \frac{(x+2)^2}{(x-3)^2-2}&\leq 0\\ \frac{(x+2)^2}{(x-3+\sqrt2)(x-3-\sqrt2)}&\leq 0 \end{aligned}$$

$$\therefore x=-2\quad\text{or}\quad 3-\sqrt2<x<3+\sqrt2$$