Quadratic Equations
Quadratic equations form the cornerstone of GCSE mathematics, appearing in over 15% of exam questions across Foundation and Higher tiers. These polynomial equations, where the highest power of x is 2, challenge Year 10 and 11 students to master factorisation, completing the square, and the quadratic formula.
Try it right now
Click “Generate a problem” to see a fresh example of this technique.
Why it matters
Quadratic equations model countless real-world scenarios that students encounter daily. A football's trajectory follows a quadratic path — if kicked from ground level at 20 metres per second, it reaches maximum height after 2 seconds using h = -5t² + 20t. Business applications include profit maximisation: a shop selling 100 items at £8 each might find that reducing price by £1 increases sales by 20 items, leading to the quadratic P = (8-x)(100+20x). Engineers use quadratics to calculate bridge spans, architects design parabolic arches, and economists model supply-demand curves. The discriminant b²-4ac determines whether solutions exist — crucial for determining if a business model is viable or if a projectile hits its target. Students who master these equations gain problem-solving skills applicable to A-level physics, economics, and advanced mathematics courses.
How to solve quadratic equations
Quadratic Equations
- Write in standard form: ax² + bx + c = 0.
- Factor, or use the quadratic formula: x = (-b ± √(b²-4ac)) / 2a.
- Check both solutions by substituting back.
Example: x² − 5x + 6 = 0 → (x−2)(x−3) = 0 → x = 2 or x = 3.
Worked examples
Find a number whose square is 25.
Answer: x = 5 or x = −5
- Set up the equation → x² = 25 — We need x such that x² = 25. This is a quadratic equation.
- Understand the equation → x² = 25 — We need to find a number that, when squared (multiplied by itself), gives us 25.
- Take the square root of both sides → x = ±√25 — When we take the square root, we must include BOTH the positive and negative root, because both (+a)² and (−a)² give a².
- Calculate √25 → √25 = 5 — Since 5 × 5 = 25, the square root of 25 is 5.
- Write both solutions → x = 5 or x = −5 — A quadratic equation can have up to 2 solutions. Here we have exactly 2.
- Verify both solutions → (5)² = 25 ✓, (−5)² = 25 ✓ — Substitute each value back into x² = 25 to confirm.
x² − 5x + 4 = 0
Answer: x = 1 or x = 4
- Write the equation in standard form → x² − 5x + 4 = 0 (a = 1, b = -5, c = 4) — Standard form is ax² + bx + c = 0. Identify a, b, and c.
- Find two numbers that multiply to c and add to b → Need: p × q = 4 and p + q = -5 → p = -4, q = -1 — We need two numbers whose product is 4 and whose sum is -5. Those are -4 and -1 because -4 × -1 = 4 and -4 + -1 = -5.
- Write the factored form → (x - 4)·(x - 1) = 0 — Rewrite the quadratic as a product of two binomials.
- Apply the zero product property → Set each factor = 0: x = 4, x = 1 — If a × b = 0, then a = 0 or b = 0. Set each factor equal to zero and solve.
- Verify by substituting back → x = 4: 4² − 5·4 + 4 = 16 − 20 + 4 = 0 ✓ — Both solutions satisfy the original equation.
Factor and solve: x² + 1x − 6 = 0
Answer: x = -3 or x = 2
- Write the equation in standard form → x² + 1x − 6 = 0 (a = 1, b = 1, c = -6) — Standard form is ax² + bx + c = 0. Identify a, b, and c.
- Find two numbers that multiply to c and add to b → Need: p × q = -6 and p + q = 1 → p = -2, q = 3 — We need two numbers whose product is -6 and whose sum is 1. Those are -2 and 3 because -2 × 3 = -6 and -2 + 3 = 1.
- Write the factored form → (x - 2)·(x + 3) = 0 — Rewrite the quadratic as a product of two binomials.
- Apply the zero product property → Set each factor = 0: x = 2, x = -3 — If a × b = 0, then a = 0 or b = 0. Set each factor equal to zero and solve.
- Verify by substituting back → x = 2: 2² + 1·2 − 6 = 4 + 2 − 6 = 0 ✓ — Both solutions satisfy the original equation.
Common mistakes
- Students often forget the negative solution when solving x² = 25, writing only x = 5 instead of x = ±5, missing half the solutions.
- When factorising x² - 7x + 12 = 0, students frequently write (x - 3)(x - 4) = 0 as x = -3 or x = -4 instead of x = 3 or x = 4, incorrectly applying the zero product property.
- Using the quadratic formula for x² + 6x + 9 = 0, students calculate x = (-6 ± 0)/2 = -3 twice but write 'x = -3 or x = -3' instead of recognising this single repeated root.
- Students mix up signs when factorising x² + x - 6 = 0, writing (x + 3)(x - 2) = 0 giving x = -3 or x = 2, but then stating 'x = 3 or x = -2' in their final answer.