WAEC 2016 · Paper 2 · Q3

Given that $P = \{x : x \in \mathbb{R},\ 2x^2 - x - 10 \le 0\}$ and $Q = \{x : x \in \mathbb{R},\ 4x^2 - 1 \ge 0\}$ , find $P \cap Q$ .

P is where the blue curve is on or below the x-axis; Q is where the red curve is on or above it. Where are both true?

Worked solution (try it first)

For $P$ : factorise $2x^2 - x - 10 = (x + 2)(2x - 5)$ .
The roots are $x = -2$ and $x = \frac52$ .
The curve is U-shaped, so it is $\le 0$ between the roots: $P = \{x : -2 \le x \le \frac52\}$ .
For $Q$ : factorise $4x^2 - 1 = (2x - 1)(2x + 1)$ .
The roots are $x = -\frac12$ and $x = \frac12$ .
It is $\ge 0$ outside the roots: $x \le -\frac12$ or $x \ge \frac12$ .
$P \cap Q$ is the part of $P$ that is also in $Q$ .
On a number line, that is from $-2$ to $-\frac12$ and from $\frac12$ to $\frac52$ , with all four ends included.
So $P \cap Q = \{x : -2 \le x \le -\frac12\} \cup \{x : \frac12 \le x \le \frac52\}$ .