📘 GCSE Maths · Notes & worksheet
Quadratics.
Notes
Three ways to solve ax² + bx + c = 0
- Factorise (when it factorises neatly). For 6x² + x − 2 find two numbers that multiply to ac = −12 and add to b = 1: that is 4 and −3. Split the middle term: 6x² + 4x − 3x − 2 = 2x(3x + 2) − 1(3x + 2) = (3x + 2)(2x − 1).
- Quadratic formula: x = (−b ± √(b² − 4ac)) / 2a. Use brackets for negative values of b.
- Completing the square: x² + bx + c = (x + b/2)² + c − b²/4.
Completing the square for a ≠ 1
Factor out a from the x terms first:
2x² + 12x + 7 = 2(x² + 6x) + 7 = 2[(x + 3)² − 9] + 7 = 2(x + 3)² − 11.
Turning points
For y = a(x + p)² + q the turning point is (−p, q). It is a minimum if a > 0 and a maximum if a < 0.
The discriminant b² − 4ac
| Value | Meaning |
|---|---|
| > 0 | two distinct real roots |
| = 0 | one repeated root (graph touches the x-axis) |
| < 0 | no real roots (graph does not cross the x-axis) |
Quadratic inequalities
- Solve the equation to find the critical values.
- Sketch the graph.
- Read off the region: above the axis for > 0, below the axis for < 0.
Example: x² − 3x − 10 > 0 gives (x − 5)(x + 2) > 0, so x < −2 or x > 5.
Algebraic fractions leading to quadratics
Multiply every term by the common denominator, expand, collect everything on one side, then solve.
Common mistakes
- Forgetting ± in the formula and giving only one solution.
- Dividing by x, which loses a solution. Factorise instead.
- Giving the turning point as (p, q) instead of (−p, q).
- Writing one inequality (−2 < x < 5) when the answer is two separate regions.
Exam tips
- If the question says "give your answer to 2 d.p." or "in surd form", use the formula.
- If it says "hence" or "write in the form (x + a)² + b", use completing the square.
Worksheet
Prints the questions only, without answers. Choose “Save as PDF” in the print window to download.
Total: 24 marks. Calculator allowed for Q2 and Q7 only.
- Solve 6x² + x − 2 = 0 by factorising. (3)
- Solve 2x² − 5x − 4 = 0. Give your answers to 2 decimal places. (3)
- Write x² − 8x + 3 in the form (x + a)² + b, and state the coordinates of the turning point of y = x² − 8x + 3. (3)
- Write 2x² + 12x + 7 in the form a(x + p)² + q. State the minimum value of the expression and the value of x where it occurs. (4)
- Find the range of values of k for which x² + kx + 9 = 0 has no real roots. (3)
- Solve the inequality x² − 3x − 10 > 0. (3)
- Solve 3/(x − 1) + 2/(x + 2) = 1. Give your answers in the form p ± √q. (5)
Show mark scheme and answers (tutor copy)
Try the worksheet first, then check your work here.
- (3x + 2)(2x − 1) = 0 (M1 attempt, A1 correct factors), so x = −2/3 or x = 1/2. (A1)
- b² − 4ac = 25 + 32 = 57. x = (5 ± √57)/4 (M1 substitution, M1 √57 ≈ 7.5498). So x = 3.14 or x = −0.64. (A1)
- (x − 4)² − 16 + 3 = (x − 4)² − 13 (M1, A1). Turning point (4, −13). (A1)
- 2(x² + 6x) + 7 (M1) = 2[(x + 3)² − 9] + 7 (M1) = 2(x + 3)² − 11 (A1). Minimum value −11 at x = −3. (A1)
- No real roots means b² − 4ac < 0: k² − 36 < 0 (M1, M1). So −6 < k < 6. (A1)
- Critical values −2 and 5 (M1). The graph is a U-shape, so the expression is positive outside the roots (M1). x < −2 or x > 5. (A1)
- 3(x + 2) + 2(x − 1) = (x − 1)(x + 2) (M1). 5x + 4 = x² + x − 2 (M1), so x² − 4x − 6 = 0 (A1). x = (4 ± √40)/2 (M1) = 2 ± √10 (A1). Check: x ≈ 5.16 gives 3/4.16 + 2/7.16 ≈ 0.721 + 0.279 = 1 ✓.