Linear & Quadratic Inequalities and Shading Graphical Regions
Inequalities compare the relative size of algebraic expressions using the symbols <, >, ≤, and ≥. Linear inequalities are solved using the same inverse operations as linear equations (with the key rule that multiplying or dividing both sides by a negative number flips the inequality sign), and can be represented on a number line or as a shaded region in the (x, y) coordinate plane. On the Higher Tier, examiners always test quadratic inequalities (ax² + bx + c > 0 or < 0), which require finding the critical values and sketching a parabola to determine whether the solution is a single connected interval or two separate tails.
- Number Line Representation & Integer Lists
- On a number line, '<' and '>' are represented by an open circle (○), whereas '≤' and '≥' are represented by a closed, filled-in circle (●). For a double inequality such as -2 < x ≤ 3, draw an open circle at -2, a closed circle at 3, and connect them with a line segment; the integers satisfying -2 < x ≤ 3 are -1, 0, 1, 2, 3 (do not include -2!).
- Solving Linear & Double Inequalities
- Use inverse operations just like an equation. For a compound/double inequality such as -5 ≤ 3x + 1 < 13, apply each inverse operation to all three parts simultaneously: subtract 1 from all three parts (-6 ≤ 3x < 12), then divide all three parts by 3 (-2 ≤ x < 4). If multiplying or dividing by a negative number, reverse the inequality signs.
- Graphical Inequalities & Shading Regions
- 1. Replace the inequality sign with '=' to get the equation of the boundary line and plot it on the grid: use a dashed line for < or > and a solid line for ≤ or ≥. 2. Substitute a test point such as (0, 0) into each inequality to confirm which side of each boundary line satisfies the inequality. 3. Label the region R that satisfies all inequalities simultaneously.
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Number Line Representation & Integer Lists
- On a number line, '<' and '>' are represented by an open circle (○), whereas '≤' and '≥' are represented by a closed, filled-in circle (●). For a double inequality such as -2 < x ≤ 3, draw an open circle at -2, a closed circle at 3, and connect them with a line segment; the integers satisfying -2 < x ≤ 3 are -1, 0, 1, 2, 3 (do not include -2!).
- Solving Linear & Double Inequalities
- Use inverse operations just like an equation. For a compound/double inequality such as -5 ≤ 3x + 1 < 13, apply each inverse operation to all three parts simultaneously: subtract 1 from all three parts (-6 ≤ 3x < 12), then divide all three parts by 3 (-2 ≤ x < 4). If multiplying or dividing by a negative number, reverse the inequality signs.
- Graphical Inequalities & Shading Regions
- 1. Replace the inequality sign with '=' to get the equation of the boundary line and plot it on the grid: use a dashed line for < or > and a solid line for ≤ or ≥. 2. Substitute a test point such as (0, 0) into each inequality to confirm which side of each boundary line satisfies the inequality. 3. Label the region R that satisfies all inequalities simultaneously.
- Quadratic Inequalities: Inside vs Outside the Roots (Higher Tier)
- To solve ax² + bx + c < 0 or > 0 (where a > 0): 1. Rearrange so one side is zero (and ensure the x² coefficient is positive). 2. Factorise (or use the quadratic formula) to find the two critical values (roots) α and β (where α < β). 3. Sketch a U-shaped parabola crossing the x-axis at α and β: if the quadratic is < 0 (or ≤ 0, below the x-axis), the solution is one combined inequality between the roots: α < x < β. If the quadratic is > 0 (or ≥ 0, above the x-axis), the solution is two separate inequalities outside the roots: x < α or x > β.
- Combining a Linear and a Quadratic Inequality (Higher Tier)
- When asked to find the set of values of x that satisfy both a linear inequality and a quadratic inequality: solve each inequality separately first, then draw both solution sets above a single number line to find where the two intervals overlap (their intersection).
- Set Notation for Inequality Intervals
- GCSE exams sometimes ask for answers in set notation: {x : x < -2} ∪ {x : x > 5} represents 'x is less than -2 OR x is greater than 5' (the union symbol ∪ means OR), while {x : -1 ≤ x ≤ 4} (or {x : x ≥ -1} ∩ {x : x ≤ 4}, where ∩ means and/intersection) represents a single continuous interval.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
List all the integers n that satisfy the inequality -3 ≤ n < 2.
Solve the inequality 11 - 2x > 5.
When graphing an inequality on an (x, y) grid, when do you draw a dashed line vs a solid line?
What are the critical values of (x + 4)(x - 2) = 0, and what is the solution to: (a) (x + 4)(x - 2) < 0, and (b) (x + 4)(x - 2) ≥ 0?
Why is writing '4 < x < -2' as the answer to x² - 2x - 8 > 0 an automatic loss of marks?
How do you find the coordinates of where the line 2x + 3y = 12 crosses the x-axis and y-axis (the cover-up method)?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Solve the double inequality -7 < 3x + 2 ≤ 17, and state the greatest integer and smallest integer that x can be. [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. Subtract 2 from all three parts of the inequality: -7 - 2 < 3x ≤ 17 - 2 -> -9 < 3x ≤ 15 [1 mark]. 2. Divide all three parts by 3: -3 < x ≤ 5 [1 mark]. 3. Since x > -3 (strictly greater) and x ≤ 5, the smallest integer x can be is -2 and the greatest integer x can be is 5 [1 mark].
- Q2
Solve the quadratic inequality 2x² - 5x - 12 > 0. Show all your algebraic steps clearly. (Higher Tier) [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Factorise the quadratic expression 2x² - 5x - 12: find two numbers that multiply to (2 x -12 = -24) and add to -5 (-8 and +3) -> (2x + 3)(x - 4) [1 mark]. 2. Set equal to zero to find the two critical values (roots): 2x + 3 = 0 -> x = -1.5 (or -3/2), and x - 4 = 0 -> x = 4 [1 mark]. 3. Sketch a U-shaped parabola crossing the x-axis at -1.5 and 4; since we require 2x² - 5x - 12 > 0, we want the two outer regions where the curve is above the x-axis [1 mark]. 4. Write the two separate inequalities: x < -1.5 or x > 4 [1 mark].
- Q3
Find the set of values of x for which both 4x - 3 > 9 - 2x and x² - 4x - 21 ≤ 0 are satisfied. (Higher Tier) [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Solve the linear inequality first: add 2x and add 3 to both sides -> 6x > 12 -> x > 2 [1 mark]. 2. Factorise the quadratic inequality x² - 4x - 21 ≤ 0: (x - 7)(x + 3) ≤ 0, giving critical values x = -3 and x = 7 [1 mark]. 3. Since the quadratic is ≤ 0 (below or on the x-axis), the quadratic solution is the single interval between the roots: -3 ≤ x ≤ 7 [1 mark]. 4. Compare both intervals on a number line to find where x > 2 and -3 ≤ x ≤ 7 overlap (intersect) [1 mark]. 5. Final combined inequality: 2 < x ≤ 7 [1 mark].
- Q4
A region R on a coordinate grid is defined by the three inequalities: x ≥ 1, y ≥ x - 1, and x + 2y < 8. Describe the boundary lines required to draw region R, and find all the points (x, y) with integer coordinates that lie inside or on the permitted boundaries of region R. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1. Boundary lines: x = 1 is a solid vertical line through x = 1; y = x - 1 is a solid diagonal line (gradient 1, y-intercept -1); and x + 2y = 8 is a dashed line connecting (0, 4) and (8, 0) [1 mark]. 2. Testing a point such as (2, 2): x = 2 ≥ 1 (True), 2 ≥ 2 - 1 (True), and 2 + 2(2) = 6 < 8 (True), so R is the triangle enclosed between the three lines (including solid lines x = 1 and y = x - 1, but excluding points on the dashed line x + 2y = 8) [1 mark]. 3. Test integer x-values systematically: For x = 1 (0 ≤ y < 3.5): (1, 0), (1, 1), (1, 2), (1, 3). For x = 2 (1 ≤ y < 3): (2, 1), (2, 2) [1 mark]. 4. For x = 3 (2 ≤ y < 2.5): (3, 2). (For x = 4, y ≥ 3 and y < 2 has no points). Total 7 integer points: (1,0), (1,1), (1,2), (1,3), (2,1), (2,2), and (3,2) [1 mark].
Common Exam Questions (4)
What is the golden rule when multiplying or dividing an inequality by a negative number?
Whenever you multiply or divide both sides of an inequality by a negative number, you must flip the direction of the inequality sign (e.g., if -3x < 12, dividing by -3 flips '<' to '>' giving x > -4)! To avoid making this mistake altogether, simply add the negative x-term to the other side first so the coefficient of x is positive.
Which circles are used on a number line, and which lines are used when shading graphical regions?
1. Strict inequalities (< or >, not equal to): use an open (empty/unshaded) circle ○ on a number line, and a dashed (dotted) boundary line on a coordinate grid. 2. Inclusive inequalities (≤ or ≥, can equal): use a closed (filled/solid) circle ● on a number line, and a solid continuous boundary line on a coordinate grid.
Why must you never solve a quadratic inequality like x² - 5x - 6 > 0 without sketching a mini U-shaped parabola? (Higher Tier)
Because a quadratic has two roots (here, critical values x = -1 and x = 6 from (x - 6)(x + 1) = 0), writing 'x > 6 and x > -1' is completely wrong and loses the final 2 marks. Sketching a quick U-shaped parabola crossing the x-axis at -1 and 6 immediately shows: 1. If the quadratic is < 0 (below the x-axis), the solution is one connected region trapped between the roots: -1 < x < 6. 2. If the quadratic is > 0 (above the x-axis), the solution is two separate regions outside the roots: x < -1 or x > 6.
How do you test which side of a diagonal line (like 2x + 3y ≤ 12) satisfies a graphical inequality?
Pick an easy test point that does not lie on the line, almost always the origin (0, 0)! Substitute x = 0 and y = 0 into the inequality: 2(0) + 3(0) ≤ 12 -> 0 ≤ 12, which is true. This proves that the side of the line containing (0, 0) is the wanted region.
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