2.6 Linear and quadratic inequalities, and set notation — Practise

Work through these on your own. Give every solution as a set: use an interval, a union, or $\mathbb{R}$ . Sketch the parabola for any quadratic inequality before you choose the region, and include or exclude each endpoint deliberately.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Solve $3x - 5 < 7$ . Give the solution in set notation.
Solve $2(x - 3) \geq 5x$ .
Solve $\dfrac{x + 1}{2} \geq 3$ .
Solve $2x - 3 < 5x + 6$ .
Solve $x^2 - x - 6 < 0$ .
Solve $x^2 \geq 4$ .
Solve $x^2 + 2x - 8 > 0$ .
Solve $x^2 - 6x + 9 \leq 0$ .
Solve $x^2 < 0$ .
Solve $3x + 1 > 3x + 5$ .
Solve $x^2 \leq 9$ .
Solve $x^2 \geq 0$ .
Solve the double inequality $-3 < 2x + 1 \leq 5$ .

Answers and concise solutions

1. The solution set

Answer: Add $5$ to both sides: $3x < 12$ . Divide by $3$ : $x < 4$ . In set notation the solution is $x \in (-\infty, 4)$ .

2. The solution set

Answer: Expand: $2x - 6 \geq 5x$ . Subtract $2x$ : $-6 \geq 3x$ . Divide by $3$ : $-2 \geq x$ , so $x \leq -2$ . In set notation, $x \in (-\infty, -2]$ . Both sides close at $-2$ because the inequality is "or equal to".

3. The solution set

Answer: Multiply both sides by $2$ : $x + 1 \geq 6$ . Subtract $1$ : $x \geq 5$ . In set notation, $x \in [5, \infty)$ .

4. The solution set

Answer: Subtract $2x$ from both sides: $-3 < 3x + 6$ . Subtract $6$ : $-9 < 3x$ . Divide by $3$ : $-3 < x$ , so $x > -3$ . In set notation, $x \in (-3, \infty)$ .

5. The solution set

Answer: Factorise: $x^2 - x - 6 = (x - 3)(x + 2)$ , so the parabola crosses the $x$ -axis at $x = -2$ and $x = 3$ . It opens upwards, so it is below the axis between the roots. The solution is $-2 < x < 3$ , or $x \in (-2, 3)$ .

6. The solution set

Answer: Rearrange to $x^2 - 4 \geq 0$ , which factorises as $(x - 2)(x + 2) \geq 0$ . The parabola opens upwards and the inequality is "greater than or equal to zero", so the solution lies outside the roots: $x \leq -2$ or $x \geq 2$ . In set notation, $x \in (-\infty, -2] \cup [2, \infty)$ .

7. The solution set

Answer: Factorise: $x^2 + 2x - 8 = (x + 4)(x - 2)$ , with roots $x = -4$ and $x = 2$ . The parabola opens upwards, so it is positive outside the roots. The inequality is strict, so the endpoints are not included: $x < -4$ or $x > 2$ , or $x \in (-\infty, -4) \cup (2, \infty)$ .

8. The solution set

Answer: Factorise: $x^2 - 6x + 9 = (x - 3)^2$ . A square is never negative, so the only way for $(x - 3)^2 \leq 0$ is $(x - 3)^2 = 0$ . The only solution is $x = 3$ .

9. The solution set

Answer: A square is never negative, so no real number satisfies $x^2 < 0$ . The solution set is empty. Compare this with $x^2 \geq 0$ , whose solution set is every real number.

10. The solution set

Answer: Subtract $3x$ from both sides: $1 > 5$ . This is false, whatever the value of $x$ , so no value of $x$ works. The solution set is empty.

11. The solution set

Answer: Rearrange to $x^2 - 9 \leq 0$ , which factorises as $(x - 3)(x + 3) \leq 0$ . The parabola opens upwards, so it is negative or zero between the roots: $-3 \leq x \leq 3$ , or $x \in [-3, 3]$ .

12. The solution set

Answer: A square is never negative, so every real number satisfies $x^2 \geq 0$ . The solution set is the whole number line, written $x \in \mathbb{R}$ .

13. The solution set

Answer: Subtract $1$ from all three parts: $-4 < 2x \leq 4$ . Divide by $2$ : $-2 < x \leq 2$ . The left end is strict and the right end is included, so $x \in (-2, 2]$ .