Which diagram best represents the solution set of $x^2 - 2x - 3 \geq 0$? A - HSC - SSCE Mathematics Extension 1 - Question 1 - 2020 - Paper 1

The intervals created by the roots are:

• $(-\infty, -1)$
• $(-1, 3)$
• $(3, \infty)$

Next, we test a point from each interval to determine where the inequality holds true:

• For $x = -2$:

$(-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5 \geq 0$ (True)

• For $x = 0$:

$0^2 - 2(0) - 3 = -3 \geq 0$ (False)

• For $x = 4$:

$4^2 - 2(4) - 3 = 16 - 8 - 3 = 5 \geq 0$ (True)

Thus, the solution set is $(-\infty, -1] \cup [3, \infty)$.

Since the solution set includes all values less than or equal to $-1$ and all values greater than or equal to $3$, we can conclude that the correct diagram is A: all values up to $-1$ and from $3$ onward.