Show that $x^2 + 4x + 9$ is always positive for all real values of $x$.[2] Hence, without u

Question

Show that $x^2 + 4x + 9$ is always positive for all real values of $x$.[2] Hence, without using a calculator, solve $\frac{(x^2 + 4x + 9)(x + 2)^2}{x^2 - 6x + 7} \le 0$.[3]

Accepted Answer

(a) $x^2+4x+9=(x+2)^2+5>0$ for all real $x$ (b) $x=-2$ or $3-\sqrt2<x<3+\sqrt2$

2024 NYJC Promo Q2

2024 NYJC Promo Q2

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