Solving quadratic inequalities algebraically and graphically, rational inequalities with linear factors, and writing solution sets in set-builder notation.
5 narrated scenes. Press play, listen and read along — the points appear as they are explained. Use ‹ Prev / Next › or the scene bar to jump to any scene.
Solving an inequality tells you where a function is positive, negative or zero. That is how you choose the right set of values — and it is a classic source of lost marks.
Narration uses your device’s built-in voice. No sound? Switch to “Captions only” — the captions advance on their own.
Copy these into your notebook in your own words — that is how they stick.
Factorise; critical values; sketch; ∪ above the axis outside the roots, below between them.
Strict: critical values excluded. ≥, ≤: included.
Critical values from numerator and denominator; sign table; denominator values are always excluded.
Unless you multiply by the square of the denominator.
Move everything to one side and combine into one fraction.
{x : −½ ≤ x ≤ 3, x ∈ ℝ}; “x < −2 or x > 3” is a union.
Test one value in each region.
Sketch the function and read where it is above/below the x-axis.
3 quick games on this lesson. Earn XP and badges as you play — your best score for each game is saved. Games are for practice and do not change your exam readiness. More games in the Additional Mathematics Arcade.
8 items modelled on real CSEC Paper 1 questions. Choose one answer for each, then check. Aim for 80% or better. Work each item on paper first.
Write full answers, then reveal the mark scheme and tick the points you earned (25 marks in total). Show all your working — method marks count. Your answers are saved on this device.
Question 1 [15 marks]
Answer all parts.
(a) Find the set of values of x for which x² − x − 6 > 0. [4]
(b) Find the set of values of x for which 2x² − 5x − 3 ≤ 0, giving your answer in set-builder notation. [4]
(c) Solve (x + 1)/(x − 2) ≥ 0, stating any value of x that must be excluded. [4]
(d) Hence, or otherwise, find the integer values of x that satisfy both 2x² − 5x − 3 ≤ 0 and (x + 1)/(x − 2) ≥ 0. [3]
Question 2 [10 marks]
Answer all parts.
(a) Show that (2x − 1)/(x + 3) < 1 is equivalent to (x − 4)/(x + 3) < 0. [3]
(b) Hence solve (2x − 1)/(x + 3) < 1. [3]
(c) Show the solution on a number line. [2]
(d) Explain why multiplying both sides of (2x − 1)/(x + 3) < 1 by (x + 3) is not a valid first step. [2]
Find these in your own past-paper booklets for extra timed practice. Items above are written in the style of these papers; any adapted from a specific question say so.