Algebra · GCSE Maths
The quadratic formula
GCSE Maths quadratic formula: write a, b and c including signs, compute the discriminant, keep the ± until the last line, and round only when the question asks.
x = [−b ± √(b² − 4ac)] / (2a). Write a, b and c from ax² + bx + c = 0, including minus signs, before you substitute a single number.
The important bits
What you need to know
- 1
Rearrange to ax² + bx + c = 0 before you name a, b and c. If the question gives x² + 5x = 6, move 6: x² + 5x − 6 = 0.
- 2
a is the coefficient of x², b of x, c the constant. For 2x² − 3x − 5 = 0, a = 2, b = −3, c = −5. The minus signs belong to b and c.
- 3
The discriminant D = b² − 4ac sits under the square root. D > 0 two real roots, D = 0 one repeated real root, D < 0 no real roots on the GCSE course.
- 4
Substitute into the formula with brackets: x = [−(−3) ± √( (−3)² − 4(2)(−5) )] / (2 × 2). −b is the first trap when b is already negative.
- 5
Keep the ± as two separate calculations after the square root is simplified. Do not divide only the square root by 2a; the whole numerator shares the denominator.
- 6
If D is a perfect square and a, b, c are integers, the roots are rational and you could have factorised. The formula still works; factorising is usually quicker.
- 7
If the question asks for 2 decimal places, compute with more decimals or leave the square root exact until the last line, then round. Rounding b² − 4ac too early shifts both roots.
- 8
Check each root in the original equation. A sign error in −b usually produces two wrong numbers that still look tidy.
Quotations worth analysing
Short evidence. Real method.
“x = (−b ± √(b² − 4ac)) / (2a)”
Write this line once from memory, then substitute. The method mark is on a, b, c and the substitution, not on a memorised pair of decimals.
“b² − 4ac”
This value decides how many real roots you should expect. A negative discriminant means stop and write “no real solutions”, not invent a calculator error.
“Give your answers to 2 decimal places”
Keep the square root unrounded until the ± split, then round each root. Rounding the discriminant to 2 d.p. first is how both answers miss the accuracy mark.
Go deeper
Signs first, then the square root
For 2x² − 3x − 5 = 0 write a = 2, b = −3, c = −5 in a tiny table. Then −b = −(−3) = 3. Discriminant = (−3)² − 4(2)(−5) = 9 + 40 = 49. The square root is 7, not ±7 yet; the ± lives in front of the 7 in the formula. So x = (3 ± 7) / 4. That is 10/4 = 2.5 or (−4)/4 = −1. Students who take b as 3 produce −b = −3 and a discriminant 9 − 40 = −31, then panic. The panic was a sign, not a quadratic that needed complex numbers. Always include the brackets around a negative b when you square it: (−3)² is 9, −3² is −9 on some calculator entries. Write (−3)² = 9 as its own line.
Go deeper
The denominator is 2a, not 2, and not a
A frequent slip is to compute −b ± √D correctly and then divide only one piece. The formula is a single fraction. For a = 2, 2a = 4, so both 3 + 7 and 3 − 7 are divided by 4. Another slip is to use 2 as the denominator whenever a = 1 is forgotten and a was actually 2. Write 2a as a number before you divide: “2a = 4”. If a is negative, 2a is negative, which is allowed; the parabola is an n-shape, and the algebra does not mind. If D is not a square, leave roots as ( −b ± √D ) / (2a) or as simplified surds unless the question asked for decimals. √12 should become 2√3 if you stay exact. Decimal answers need the degree of accuracy printed on the paper, usually 2 d.p. or 3 s.f.
Go deeper
Discriminant as a planning tool
Before grinding the formula, glance at D. If D is 0, there is one root, x = −b / (2a), the turning point on the x-axis. If D is 1, 4, 9, 16, 25, 36, 49, factorising is probably faster and you can use the formula as a check. If D is 13, you will need surds or decimals; do not waste minutes hunting integer pairs. If D is negative, write “no real roots” and move on — Higher papers still award that statement. The discriminant also explains the graph: D > 0 crosses the x-axis twice, D = 0 touches once, D < 0 misses. A later part that asks for the turning point is completing the square, not a second run of the formula, unless you use x = −b / (2a) for the axis of symmetry and substitute back for y.
See the idea in action
Solve 2x² − 3x − 5 = 0. Give exact answers. Step 1: a = 2, b = −3, c = −5. Step 2: Discriminant b² − 4ac = (−3)² − 4(2)(−5) = 9 + 40 = 49. Step 3: √49 = 7. Step 4: x = [−(−3) ± 7] / (2 × 2) = (3 ± 7) / 4. Step 5: x = (3 + 7)/4 = 10/4 = 5/2, or x = (3 − 7)/4 = −4/4 = −1. Check: 2(5/2)² − 3(5/2) − 5 = 2(25/4) − 15/2 − 5 = 25/2 − 15/2 − 10/2 = 0. Check: 2(1) − 3(−1) − 5 = 2 + 3 − 5 = 0.
Exam technique
Turn knowledge into marks
Write a, b and c including signs as a labelled list, then the discriminant, then the formula with brackets. If the paper asks for 2 d.p., round only on the last line.
Common mistakes
Do not give these marks away
- 01
Taking b as positive when it is negative, so −b and b² − 4ac are both wrong.
- 02
Dividing only the square root by 2a, or using denominator 2 when a is not 1.
- 03
Rounding the discriminant or the square root too early, then missing the 2 decimal place answers.
For 2x² − 3x − 5 = 0, the discriminant b² − 4ac is
A49
B−31
C9
D40
Show the answer
49. b = −3 and c = −5, so (−3)² − 4(2)(−5) = 9 + 40 = 49. −31 comes from taking 4ac as positive. 9 is b² alone. 40 is 4ac without b².
Quick questions
If this is the bit you searched
When must I use the quadratic formula?
When the quadratic does not factorise over the integers, or when the question asks for decimal roots. Factorise when the numbers are friendly; the formula always works.
How do I remember the quadratic formula?
Say “minus b, plus or minus the square root of b squared minus 4ac, all over 2a.” Write a, b, c first so the substitution is mechanical.
What does a negative discriminant mean at GCSE?
No real roots. The graph of y = ax² + bx + c does not meet the x-axis. You are not required to give complex solutions.
Why did I get one correct root and one wrong one?
Usually the ± was applied to only part of the numerator, or 2a was used on one branch only. Compute ( −b + √D ) / (2a) and ( −b − √D ) / (2a) as two full fractions.