OxfordAQA International GCSE Mathematics: Solving quadratics: factorising, completing the square (9260) – Revision Notes

These revision notes condense reference A20 in section 3.2.3 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). Factorising and graph solutions are Core. Completing the square and the quadratic formula are marked “Extension only”. The specification says all content can be assessed on the Extension tier question papers, so Core content is also Extension content. For full explanations, read the study guide.

Test yourself with the practice questions. The course page is OxfordAQA IGCSE Mathematics, the printable list is the checklist, and the free 10-minute diagnostics show where to start.

Key facts

Item	Fact
Standard form	ax² + bx + c = 0, with a ≠ 0
Zero-product rule	When p × q = 0, p = 0 or q = 0. This needs the product to be exactly 0
Solutions	The x-values where y = ax² + bx + c meets y = 0
Number of solutions	Two, one (repeated) or none
Quadratic formula (Extension only)	x = (−b ± √(b² − 4ac)) / (2a)
Completing the square (Extension only)	x² + bx + c = (x + b/2)² − (b/2)² + c

Method in steps: factorising (Core)

1. Expand brackets and collect terms so the equation is “= 0”.
2. Look for a common factor, then a difference of two squares.
3. Otherwise factorise the trinomial.
4. Set each bracket to 0 and solve.
5. Check one solution by substituting.

Trinomial patterns

Type	Method	Example
x² + bx + c	A pair with product c and sum b	x² − 7x + 12 = (x − 3)(x − 4)
ax² + bx + c	A pair with product ac and sum b; split the middle, factorise in pairs	3x² + 11x − 4 = (x + 4)(3x − 1)
Common factor	Take it out first	2x² − 8x = 2x(x − 4)
a² − b²	(a − b)(a + b)	x² − 81 = (x − 9)(x + 9)
Perfect square	One repeated factor	x² − 10x + 25 = (x − 5)²

Small worked reminders

Solve 3x² + 11x − 4 = 0. Here ac = −12 and the numbers 12 and −1 have product −12 and sum 11. So 3x² + 12x − x − 4 = 3x(x + 4) − (x + 4) = (x + 4)(3x − 1). Then x = −4 or x = 1/3.

Solve 2x² − 8x = 0. Then 2x(x − 4) = 0, so x = 0 or x = 4. Do not divide by x.

Solve x² − 81 = 0. Then (x − 9)(x + 9) = 0, so x = 9 or x = −9.

Method in steps: graphs (Core)

1. To solve ax² + bx + c = 0 from the graph of y = ax² + bx + c, read the x-values where the curve crosses the x-axis.
2. To solve a different equation, rearrange it until one side equals the curve’s expression. The other side gives the line to draw.
3. Draw that line (a horizontal line is y = number). Read the x-values where it meets the curve.
4. Give answers to the accuracy asked for and describe them as approximate.

Example. The graph of y = x² − 4x − 1 is drawn. To solve x² − 4x − 3 = 0, write x² − 4x − 1 = 2. Draw y = 2. The line meets the curve at about x = −0.6 and x = 4.6.

Method in steps: completing the square (Extension only)

1. Write x² + bx + c with the x² coefficient equal to 1. If a ≠ 1, take a out of the x-terms only.
2. Halve b. Write (x + b/2)² and subtract (b/2)².
3. Add the constant term.
4. Rearrange to (x + b/2)² = a number, take the square root with ±, then subtract b/2.

Example. x² − 6x + 2 = 0. Then (x − 3)² − 9 + 2 = (x − 3)² − 7 = 0. So x = 3 ± √7, which is 5.65 or 0.35 to 2 decimal places.

Method in steps: formula (Extension only)

1. Write the equation as “= 0” and note a, b and c with signs.
2. Substitute using brackets.
3. Evaluate b² − 4ac.
4. If it is negative, there are no real solutions. Otherwise finish both ± values.

Example. 2x² + 5x − 1 = 0. Here b² − 4ac = 25 + 8 = 33, so x = (−5 ± √33)/4, which is 0.186 or −2.69 to 3 significant figures.

Choosing the line on a graph

If the curve drawn is y = x² − 4x − 1, rewrite the equation you want so that the left side is x² − 4x − 1. The right side is the line.

Equation to solve	Rewritten	Line to draw
x² − 4x − 1 = 0	already matches	y = 0 (the x-axis)
x² − 4x − 3 = 0	x² − 4x − 1 = 2	y = 2
x² − 4x = 0	x² − 4x − 1 = −1	y = −1
x² − 4x + 2 = 0	x² − 4x − 1 = −3	y = −3

To find the line, subtract the extra constant from both sides until the left side matches the curve. Then check: rewrite your line as an equation and make sure it returns to the one you were given. If it does not, the line is wrong.

Equations from a context

Many questions give a shape or a situation, ask you to form an equation, then solve it.

1. Define the unknown with a letter and units.
2. Write the relationship, for example length × width = area.
3. Expand and rearrange to “= 0”.
4. Solve, then reject any solution that does not fit, such as a negative length.
5. Answer in the words of the question, with units.

Example. A rectangle has sides (x + 3) cm and x cm and area 40 cm². Then x(x + 3) = 40, so x² + 3x − 40 = 0 and (x + 8)(x − 5) = 0. The side x must be positive, so x = 5. The sides are 5 cm and 8 cm. (Check: 5 × 8 = 40.)

Accuracy and presentation

• If the question says “give your answer to 3 significant figures”, the exact form is not enough. Write the rounded values.
• If the question says “leave your answer in surd form”, do not convert to decimals.
• If it says “show that”, every line of algebra must appear, including the line where the equation reaches its final form.
• Put the two solutions together in the final line, for example x = −8 or x = 6, so the answer is easy to find.

Must-know distinctions

• Solving an equation is different from factorising an expression. Solutions are values of x; a factorised form is not a solution.
• A root, a solution and an x-intercept refer to the same x-values here.
• Exact form (a surd) and decimal form are different. Follow the question’s instruction on accuracy.
• The formula gives two values from one calculation. Do not drop the minus.
• A graph gives approximate values. Algebra gives exact values.

Quick self-test

1. Solve x² − x − 56 = 0.
2. Solve x² − 3x = 0.
3. Solve 4x² − 49 = 0.
4. Solve 2x² + 7x − 4 = 0.
5. Rearrange and solve x(x − 1) = 12.
6. A graph of y = x² − 4x + 1 is drawn. Which line do you draw to solve x² − 4x − 2 = 0?
7. (Extension only) Write x² + 6x − 5 in the form (x + p)² + q.
8. (Extension only) Solve x² − 4x + 1 = 0, leaving the answer in surd form.
9. (Extension only) Use the formula to solve 2x² + 5x − 1 = 0 to 3 significant figures. State b² − 4ac first.
10. (Extension only) Show that x² − 3x + 5 = 0 has no real solutions.

Answers

1. (x − 8)(x + 7) = 0, so x = 8 or x = −7.
2. x(x − 3) = 0, so x = 0 or x = 3.
3. (2x − 7)(2x + 7) = 0, so x = 3.5 or x = −3.5.
4. (2x − 1)(x + 4) = 0, so x = 1/2 or x = −4.
5. x² − x − 12 = 0, so (x − 4)(x + 3) = 0 and x = 4 or x = −3.
6. x² − 4x − 2 = 0 means x² − 4x + 1 = 3. Draw y = 3.
7. (x + 3)² − 9 − 5 = (x + 3)² − 14.
8. (x − 2)² − 4 + 1 = 0 gives (x − 2)² = 3, so x = 2 ± √3.
9. b² − 4ac = 25 + 8 = 33, so x = (−5 ± √33)/4, which is 0.186 or −2.69.
10. b² − 4ac = 9 − 20 = −11, which is negative, so there are no real solutions.

Where marks are usually lost

• Not rearranging to “= 0” before factorising, for example solving (x + 1)(x + 4) = 10 by setting each bracket to 10.
• Dividing by x and losing the solution x = 0.
• Splitting the middle term with the wrong signs when a ≠ 1.
• Stopping at the factors and not stating the values of x.
• Reading y-values from the graph when the question asks for x.
• Drawing the wrong line for a rearranged equation, such as y = −2 when y = 2 is needed.
• Halving the wrong number when completing the square with a ≠ 1.
• Writing −b² instead of (−b)² in the formula when b is negative.
• Rounding in the middle of the working and then giving an inaccurate final answer.
• Giving a decimal when the question asks for surd form, or the reverse.

Examiners can only give credit for what is written. Showing the factorised form, the substitution into the formula or the completed-square form makes your method clear even if a later slip occurs.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.2.3, reference A20.