Key Stage 4 maths · Algebra
GCSE higher: quadratic formula, iteration and inequalities
This is the equation-solving topic for pupils working at the top of GCSE maths. They solve quadratics that do not factorise by completing the square or using the quadratic formula, solve simultaneous equations when one is quadratic, find approximate solutions by iteration, and solve linear and quadratic inequalities, showing the answers on a graph or in set notation.
Quadratic formula, iteration, inequalities, explained
Many quadratic equations will not factorise neatly, so two other methods are taught. Completing the square rewrites x² + 6x − 4 = 0 as (x + 3)² − 13 = 0, so x + 3 = ±√13 and x = −3 ± √13. The quadratic formula, x = (−b ± √(b² − 4ac)) ÷ 2a, solves any equation written as ax² + bx + c = 0. Some equations need rearranging first, such as x² + 3x = 4 − x, which must become x² + 4x − 4 = 0 before a, b and c can be read off. Answers are given as exact surds or rounded as the question asks.
In simultaneous equations where one equation is linear and the other quadratic, the usual approach is substitution. Take y = x + 1 and x² + y² = 25. Replace y in the second equation to get x² + (x + 1)² = 25, which simplifies to x² + x − 12 = 0, so (x + 4)(x − 3) = 0. That gives x = −4 or x = 3, and the linear equation then gives y = −3 or y = 4. The solutions must be written as pairs, here (−4, −3) and (3, 4). Graphically, these are the two points where a straight line crosses a circle.
Iteration finds approximate solutions by repeating a calculation. Rearrange x³ − 3x − 5 = 0 into x = ∛(3x + 5), start with x₀ = 2, and feed each answer back in. To two decimal places the values run 2.22, 2.27, 2.28, 2.28 and settle near 2.28. Pupils use the calculator’s answer key to repeat the step quickly. To show that a root is 2.28 to two decimal places, they can also show a change of sign: the expression is negative at 2.275 and positive at 2.285, so a root lies between.
A linear inequality in two variables, such as y > 2x − 1, is shown by drawing the boundary line and shading a region. The line is dashed when the inequality is strict (< or >) and solid when it includes equality (≤ or ≥). Testing a point such as (0, 0) shows which side is correct. Several inequalities together mark out a region, and schools differ on whether to shade the wanted region or the unwanted one, so follow the instruction on the question.
A quadratic inequality in one variable is solved by finding where the quadratic equals zero and then sketching the curve. To solve x² − x − 6 < 0, factorise to (x − 3)(x + 2), so the curve crosses the axis at −2 and 3. It is below the axis between them, so −2 < x < 3. For x² > 9 the answer has two parts, x < −3 or x > 3. Set notation writes the solution as a set, for example {x : −2 < x < 3}, meaning “the set of all x such that x is between −2 and 3”.
Words your child will hear
- Completing the square
- Rewriting x² + bx + c as (x + b/2)² plus or minus a number, to solve the equation or find the turning point.
- Quadratic formula
- A formula that gives the solutions of ax² + bx + c = 0 from the numbers a, b and c.
- Surd
- A root that cannot be written exactly as a whole number or fraction, such as √13. Leaving it in surd form keeps the answer exact.
- Iteration
- Repeating a rule, using each answer as the next input, to home in on a solution.
- Strict inequality
- One that uses < or > and does not include the boundary value.
- Boundary line
- The line, such as y = 2x − 1, that separates the region that fits an inequality from the rest of the graph.
- Set notation
- A way of writing a solution as a set, such as {x : x < −3}, read as “the set of x such that x is less than −3”.
What the National Curriculum says
In Key Stage 4, pupils should be taught to:
- Solve quadratic equations, including those that require rearrangement, algebraically by completing the square and by using the quadratic formula
- Solve 2 simultaneous equations in 2 variables (linear/quadratic) algebraically
- Find approximate solutions to equations numerically using iteration
- Solve linear inequalities in 2 variables and represent the solution set on a graph
- Solve quadratic inequalities in 1 variable
- Represent the solution set of an inequality using set notation
Objectives from the National Curriculum in England, © Crown copyright, used under the Open Government Licence v3.0.
A worked example
Solve 2x² + 3x − 4 = 0 using the quadratic formula. Give your answers to 2 decimal places.
- The equation already equals zero. Read off the numbers in front: a = 2, b = 3 and c = −4.
- Write the formula: x = (−b ± √(b² − 4ac)) ÷ 2a.
- Substitute, using brackets for the negative number so no sign is lost.
x = (-3 ± √(3² - 4×2×(-4))) ÷ (2×2)
- Work out inside the root: 9 − (−32) = 9 + 32 = 41. Subtracting a negative adds. The expression is x = (−3 ± √41) ÷ 4, which is the exact answer.
- Evaluate each version. √41 is about 6.4031.
x = (-3 + 6.4031) ÷ 4 = 0.85 x = (-3 - 6.4031) ÷ 4 = -2.35
x = (−3 ± √41) ÷ 4, so x = 0.85 or x = −2.35 to 2 decimal places.
Common mistakes and how to help
Sign slips in the formula, especially when b or c is negative
Negative numbers are squared and subtracted in the same line. For b = −5, “−b” is +5, and when a is positive and c = −4, the term −4ac becomes positive.
What helps: Always put substituted numbers in brackets and type the whole thing into the calculator carefully. Check by putting the answers back into the original equation.
Giving one solution, or only half of a quadratic inequality
Solving x² > 9 by taking a square root gives x > 3 and the other region is forgotten. The ± is also often dropped in the formula.
What helps: Sketch the parabola. Wherever the curve is above the axis is x² − 9 > 0, which is on both sides, giving x < −3 or x > 3.
Getting the number after completing the square wrong
Expanding (x + 3)² gives 9 as well as 6x, so the 9 has to be taken off again. Pupils write (x + 3)² − 4 instead of (x + 3)² − 13.
What helps: Expand the answer to check. (x + 3)² − 13 = x² + 6x + 9 − 13 = x² + 6x − 4, which is the original.
Rounding too early in iteration, or stopping at the first answer
Rounding every step to one or two decimal places can send the later values off course, and the first answer feels like the solution.
What helps: Keep every digit on the calculator and round only the final answer. Stop when successive values agree to the accuracy asked for.
How to help at home
- Put the answers back in. After solving any equation, substitute each solution into the original. For the worked example, 2 × 0.85² + 3 × 0.85 − 4 comes out very close to 0. This habit catches sign slips.
- Try a calculator iteration. Type any positive number, then repeat “add 2, then square root” using the answer key. The results settle at 2. Ask why: 2 is the number where √(2 + 2) = 2 again.
- Turn a budget into an inequality. Pears cost 60p and apples 40p, and you can spend at most £3. With x pears and y apples, working in pence, that is 60x + 40y ≤ 300. List some combinations that fit. On a graph, the whole-number points inside the region are all the possible shopping baskets.
- Test points on a number line. For x² − x − 6 < 0, try x = −3, 0 and 4 in the expression and see which ones work. It shows why the answer is a single range, and the habit helps on every inequality.
- Explain which method and why. Give your teenager three quadratics and ask which method they would choose for each, and why. Spotting the quickest method is a real skill.
Related topics
Builds on
- Key Stage 3: Algebraic notation and formulae
- Key Stage 3: Solving linear equations
- Key Stage 3: Graphs of linear and quadratic functions
- Key Stage 3: Sequences and the nth term
More on algebra
- Key Stage 4: Straight-line and quadratic graphs
- Key Stage 4: Quadratics, simultaneous equations, inequalities
- Key Stage 4: Special sequences and the nth term
- Key Stage 4: Algebraic fractions, proof and functions
- Key Stage 4: Exponential, trig and circle graphs
- Key Stage 4: Quadratic sequences and surd ratios
- Key Stage 4: Factorising, indices and identities
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