Quadratic Equations and Expressions
Quadratics appear somewhere in almost every IGCSE Maths paper. There are three ways to solve them, and a large part of the skill is knowing which one the question is asking for — because choosing the wrong method can cost you every mark even when your answer is right.
1. What counts as a quadratic
A quadratic is an expression or equation in which the highest power of the variable is 2.
The general form is:
ax² + bx + c = 0, where a ≠ 0
| Equation | Quadratic? | Why |
|---|---|---|
| x² + 5x + 6 = 0 | Yes | highest power is 2 |
| x² + 3 = 0 | Yes | b is simply 0 |
| x² = 0 | Yes | b and c are both 0 |
| 3x + 1 = 0 | No | highest power is 1 — linear |
| x³ + x² = 0 | No | highest power is 3 — cubic |
A missing middle term does not stop it being quadratic. Equations like x² + 3 = 0 or x² − 16 = 0 are quadratics with b = 0. The only requirement is an x² term with a ≠ 0.
2. Expanding brackets — get this right first
Nearly every mistake later in the topic traces back to expanding a squared bracket wrongly. The two errors below are the most common in the whole topic:
(x + 1)² is NOT x² + 1 (x + 5)² is NOT x² + 25
(x + a)² means (x + a)(x + a) — you must multiply it out properly:
(x + a)² = x² + 2ax + a² (x − a)² = x² − 2ax + a²
Worked check:
- (x + 1)² = x² + 2x + 1
- (x + 5)² = x² + 10x + 25
- (x − 3)² = x² − 6x + 9
- (2x − 4)² = 4x² − 16x + 16 ← note the 4x², not x²
When the bracket starts with 2x, the first term is 4x². Forgetting to square the coefficient as well as the letter is a frequent slip.
3. Factorising quadratics
Step 0 — always look for a common factor first
3x² − 12x = 3x(x − 4)
Take out everything that is common — number and letter. Half-factorising (taking out 3 but not x) loses marks and makes the next step harder.
Type 1 — x² + bx + c (when a = 1)
Find two numbers that multiply to c and add to b.
Example: x² + 7x + 12
- multiply to 12, add to 7 → 3 and 4
- (x + 3)(x + 4)
Example: x² − 5x + 6
- multiply to +6, add to −5 → −2 and −3
- (x − 2)(x − 3)
Signs decide the search. If c is positive, both numbers have the same sign (both negative if b is negative). If c is negative, the numbers have opposite signs. Working this out first halves the number of pairs you have to try.
Check by expanding. If your two brackets don’t multiply back to the original, you have made an error — this takes ten seconds and catches most mistakes.
Type 2 — difference of two squares
a² − b² = (a + b)(a − b)
- x² − 16 = (x + 4)(x − 4)
- 9x² − 25 = (3x + 5)(3x − 5)
a² − b² is not (a − b)². These get confused constantly. The difference of two squares gives two different brackets; (a − b)² is one bracket squared, which expands to a² − 2ab + b².
This only works for a subtraction. x² + 16 does not factorise.
Type 3 — ax² + bx + c (when a ≠ 1)
Use splitting the middle term:
- Multiply a × c
- Find two numbers that multiply to ac and add to b
- Split the middle term into those two terms
- Factorise in pairs
Example: 3x² + 10x + 8
- a × c = 3 × 8 = 24
- multiply to 24, add to 10 → 6 and 4
- 3x² + 6x + 4x + 8
- 3x(x + 2) + 4(x + 2) = (3x + 4)(x + 2)
The two brackets in step 4 must be identical. If they aren’t, something earlier is wrong — usually the pair of numbers or a sign. That check is the fastest error-detector in this topic.
4. Solving by factorisation
Set the equation to zero first. Always.
If two things multiply to give zero, at least one of them must be zero.
Example: solve x² + 7x + 12 = 0
- Factorise: (x + 3)(x + 4) = 0
- Set each bracket to zero: x + 3 = 0 or x + 4 = 0
- Solve: x = −3 or x = −4
The signs flip. (x + 3) = 0 gives x = −3, not +3. This single step accounts for a large share of lost marks.
Example: solve x² = 5x
Do not divide both sides by x — that destroys a solution.
- x² − 5x = 0
- x(x − 5) = 0
- x = 0 or x = 5
Dividing by x loses the root x = 0. Move everything to one side and factorise instead.
A quadratic normally has two solutions — give both unless the question restricts them.
5. The quadratic formula
For ax² + bx + c = 0:
x = [ −b ± √(b² − 4ac) ] / (2a)
Example: solve 2x² + 3x − 7 = 0 (3 s.f.)
- a = 2, b = 3, c = −7
- b² − 4ac = 9 − 4(2)(−7) = 9 + 56 = 65
- x = (−3 ± √65) / 4
- √65 ≈ 8.0623
- x = (−3 + 8.0623)/4 = 1.27 or x = (−3 − 8.0623)/4 = −2.77
The things that go wrong, in order of frequency:
Losing the ±. It produces both answers. Without it you can score at most half the marks.
Sign errors in −4ac. With c negative, −4ac becomes positive — that’s why 9 − 4(2)(−7) gave 9 + 56. Students very often compute 9 − 56 and then try to square-root a negative.
Calculator input. Put brackets round negative numbers: entering −3² gives −9 on most calculators, but (−3)² gives 9. Several marks per paper are lost to exactly this.
A negative discriminant means no real solutions. If b² − 4ac comes out negative in a 0580 question, you have almost certainly made an arithmetic slip — check before writing “no solution”.
The quadratic formula is normally provided on the Extended formula list, but the method for completing the square is not. Check your own paper’s formula sheet so you know what you must memorise.
6. Completing the square
Writing a quadratic as (x + p)² + q.
The idea
Take x² + 6x. We want (x + 3)², but expanding that gives x² + 6x + 9 — a 9 we didn’t ask for. So we subtract it back:
x² + 6x = (x + 3)² − 9
That’s the whole method: halve the middle coefficient, square the bracket, then subtract the extra constant.
When a = 1
Example: write x² − 2x − 4 in the form (x − a)² + b
- Halve the middle coefficient: −2 ÷ 2 = −1
- Write (x − 1)² — which expands to x² − 2x + 1
- Subtract the extra 1, and keep the −4: x² − 2x − 4 = (x − 1)² − 1 − 4
- (x − 1)² − 5
Example: x² + 8x + 6
- half of 8 = 4 → (x + 4)² = x² + 8x + 16
- (x + 4)² − 16 + 6 = (x + 4)² − 10
You subtract the square of the halved number, not the original. With x² + 8x you subtract 16, not 8.
When the coefficient of x is a fraction
Example: x² − 3x + 7
- half of −3 = −3/2 → (x − 3/2)² = x² − 3x + 9/4
- x² − 3x + 7 = (x − 3/2)² − 9/4 + 7
- −9/4 + 7 = −9/4 + 28/4 = 19/4
- (x − 3/2)² + 19/4
Fractions are normal here and are not a sign you’ve gone wrong. Keep them as fractions — converting to decimals early loses accuracy and marks.
When a ≠ 1
Factor out the coefficient of x² first, from the x² and x terms only.
Example: 3x² − 18x − 3
- Factor 3 from the first two terms: 3(x² − 6x) − 3
- Complete the square inside: x² − 6x = (x − 3)² − 9
- 3[(x − 3)² − 9] − 3
- Expand the 3 across both terms: 3(x − 3)² − 27 − 3
- 3(x − 3)² − 30
Step 4 is where marks are lost. The 3 multiplies the −9 as well, giving −27. Forgetting this is the single most common error in a ≠ 1 questions.
Reading the turning point
For y = (x + p)² + q, the turning point is at (−p, q).
- y = (x − 1)² − 5 → turning point (1, −5), a minimum
- y = (x + 4)² − 10 → turning point (−4, −10), a minimum
The x-coordinate has the opposite sign to the number in the bracket. (x − 1)² gives x = +1.
If the coefficient of x² is positive the parabola opens upward (minimum); if negative, it opens downward (maximum). The line of symmetry is the vertical line through the turning point, x = −p.
Solving by completing the square
Solve x² − 2x − 4 = 0
- (x − 1)² − 5 = 0
- (x − 1)² = 5
- x − 1 = ±√5 ← the ± is essential
- x = 1 ± √5, so x = 3.24 or x = −1.24 (3 s.f.)
Square-rooting gives two answers. Writing x − 1 = √5 alone loses half the solution.
“Complete the square” and “hence solve” are two instructions. If the question says hence, you must go on and solve — stopping at the completed-square form leaves marks behind.
7. Forming quadratics from word problems
Typical shapes: consecutive integers, areas of rectangles, ages, or a length expressed in terms of x.
Method:
- Define the variable clearly — “let the width be x cm”
- Write the relationship the question describes
- Expand and rearrange to = 0
- Solve
- Check the answers against the context
Reject impossible solutions, and say that you have. A length, age, or number of people cannot be negative. If you get x = 8 or x = −3 for a width, write “x = −3 is rejected as a length cannot be negative” and continue with x = 8. Simply omitting it can cost the reasoning mark.
“Show that…” questions give you the answer. Your job is the working, and every step must be visible — you cannot earn those marks by asserting the result.
8. Which method should you use?
This decides marks more often than the algebra does.
| The question says | What you must do |
|---|---|
| ”Factorise” | Factorise — the formula scores nothing |
| ”Solve by factorisation” | Factorise, then solve |
| ”Complete the square” | Completing the square only |
| ”Solve” (no method named) | Any valid method |
| ”Give your answer to 2 d.p.” | Expect the formula — it rarely factorises |
| ”Give an exact answer” | Leave surds in, e.g. 1 ± √5 — don’t convert to decimals |
If a method is named, using another one scores zero even with a correct answer. Read the command word before you start.
Answers to 2 or 3 significant figures are a strong hint that the quadratic does not factorise, so go straight to the formula rather than hunting for factors that don’t exist.
On a non-calculator paper, the numbers will factorise or complete neatly. If they aren’t behaving, re-check your rearrangement.
9. Mistakes that cost marks
Expanding (x + a)² as x² + a². It is x² + 2ax + a².
Not setting the equation to zero before factorising.
Forgetting the ± in the formula or when square-rooting.
Sign errors in −4ac when c is negative.
Dividing through by x, destroying the root x = 0.
Reading roots with the wrong sign — (x + 3) = 0 gives x = −3.
Confusing a² − b² with (a − b)².
Not factoring out a before completing the square, or forgetting to multiply the constant back.
Subtracting the halved number instead of its square.
Giving only one solution when the question wants both.
Keeping a negative answer for a length, age or count.
Using the formula when the question said “factorise”.
Rounding too early — carry full accuracy through and round only at the end.
Frequently asked questions
What is a quadratic equation? An equation in which the highest power of the unknown is 2, of the form ax² + bx + c = 0 with a ≠ 0.
What are the three ways to solve a quadratic? Factorising, the quadratic formula, and completing the square. Use whichever the question asks for; if it doesn’t specify, any of them is acceptable.
What is the quadratic formula? x = [−b ± √(b² − 4ac)] / (2a)
How do you complete the square? Halve the coefficient of x, write the bracket squared, then subtract the square of that halved number and simplify. If the coefficient of x² isn’t 1, factor it out first.
How do you find the turning point of a quadratic? Write it as (x + p)² + q — the turning point is (−p, q).
Why do I get two answers? A quadratic graph normally crosses the x-axis twice. Both values satisfy the equation, so give both unless the context rules one out.
When can I ignore a solution? When it is impossible in context — a negative length, age or quantity. Say explicitly that you are rejecting it and why.
What does it mean if b² − 4ac is negative? There are no real solutions. In a 0580 question this usually means an arithmetic error — check your signs first.
How do I factorise when the x² coefficient isn’t 1? Multiply a × c, find two numbers multiplying to that and adding to b, split the middle term, then factorise in pairs.
Quick revision checklist
- I can recognise a quadratic, including when b or c is 0
- I can expand (x + a)² correctly every time
- I always take out a common factor first
- I can factorise x² + bx + c using the multiply/add pairs
- I use the signs of b and c to narrow the search
- I can spot and factorise a difference of two squares
- I can factorise when a ≠ 1 by splitting the middle term
- I always set the equation to zero before solving
- I read the roots off with the correct signs
- I never divide through by x
- I can use the quadratic formula, including the ± and negative c
- I put brackets round negatives on the calculator
- I can complete the square when a = 1, with fractions, and when a ≠ 1
- I remember to multiply the constant back when I factor out a
- I can read the turning point and line of symmetry from completed-square form
- I can form a quadratic from a word problem and define my variable
- I reject impossible solutions and say so
- I check the command word before choosing a method
These notes cover quadratic equations and expressions in the Cambridge IGCSE Mathematics (0580) syllabus, and are written for Grade 9–11 / Year 10–11 students. They are based on teaching patterns observed across a large set of one-to-one IGCSE Maths lessons — this was the single most-taught topic in that set — with particular attention to the errors students make most often and the wording examiners reward. Always check the current syllabus and formula list for your own exam series.
Finished this topic?
Saved on this device — no account needed.