Quadratic Equations
A quadratic equation is a polynomial equation of degree 2, written in the standard form ax² + bx + c = 0, where a ≠ 0. These equations produce curved graphs called parabolas and can have 0, 1, or 2 real solutions. The solutions represent the x-intercepts where the parabola crosses the horizontal axis.
Why it matters
Quadratic equations model countless real-world situations involving area, projectile motion, and optimisation problems. Engineers use them to calculate the trajectory of a football kicked 15 metres across a pitch, whilst architects apply quadratic principles when designing curved structures like bridges spanning 100-metre rivers. In business, quadratic models help determine maximum profit when selling 250 concert tickets at £20 each. Physics students encounter quadratics when calculating how long a ball takes to fall 45 metres from a building. The discriminant (b² - 4ac) predicts whether a quadratic has 2, 1, or 0 real solutions, essential knowledge for GCSE students tackling Year 10 factoring problems and Year 11 completing the square techniques. These skills directly support A-level mathematics, where quadratics appear in calculus, trigonometry, and complex number theory.
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
A square has an area of 9 cm². What is the side length?
Answer: x = 3 cm (side length must be positive)
- Interpret the context → Side length = x, so x² = 9 — A square's area equals side length squared. Since length must be positive, we only keep the positive root.
- Understand the equation → x² = 9 — We need to find a number that, when squared (multiplied by itself), gives us 9.
- Take the square root of both sides → x = ±√9 — 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 √9 → √9 = 3 — Since 3 × 3 = 9, the square root of 9 is 3.
- Write both solutions → x = 3 or x = −3 — A quadratic equation can have up to 2 solutions. Here we have exactly 2.
- Verify both solutions → (3)² = 9 ✓, (−3)² = 9 ✓ — Substitute each value back into x² = 9 to confirm.
A rectangle's length is 1 cm more than its width. Its area is 20 cm². Find the dimensions.
Answer: Width = 4 cm, Length = 5 cm
- Set up the equation → Let x = width. Then length = x + 1, area = x(x + 1) = 20 → x² + 1x − 20 = 0 — Width is x, length is x + 1. Area = width × length = 20.
- Write the equation in standard form → x² − 9x + 20 = 0 (a = 1, b = -9, c = 20) — 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 = 20 and p + q = -9 → p = -5, q = -4 — We need two numbers whose product is 20 and whose sum is -9. Those are -5 and -4 because -5 × -4 = 20 and -5 + -4 = -9.
- Write the factored form → (x - 5)·(x - 4) = 0 — Rewrite the quadratic as a product of two binomials.
- Apply the zero product property → Set each factor = 0: x = 5, x = 4 — If a × b = 0, then a = 0 or b = 0. Set each factor equal to zero and solve.
- Verify by substituting back → x = 5: 5² − 9·5 + 20 = 25 − 45 + 20 = 0 ✓ — Both solutions satisfy the original equation.
x² − 3x + 0 = 0
Answer: x = 0 or x = 3
- Write the equation in standard form → x² − 3x + 0 = 0 (a = 1, b = -3, c = 0) — 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 = 0 and p + q = -3 → p = 0, q = -3 — We need two numbers whose product is 0 and whose sum is -3. Those are 0 and -3 because 0 × -3 = 0 and 0 + -3 = -3.
- Write the factored form → x·(x - 3) = 0 — Rewrite the quadratic as a product of two binomials.
- Apply the zero product property → Set each factor = 0: x = 0, 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 = 0: 0² − 3·0 + 0 = 0 + 0 + 0 = 0 ✓ — Both solutions satisfy the original equation.
Common mistakes
- Forgetting to write equations in standard form leads to incorrect identification of coefficients, such as treating 3x² - 7 = 2x as having b = 2 instead of b = -2 after rearranging to 3x² - 2x - 7 = 0
- Applying the zero product property incorrectly produces wrong solutions, like solving (x - 3)(x + 5) = 0 as x = 3 and x = 5 instead of x = 3 and x = -5
- Calculation errors with the discriminant create false conclusions about solution types, such as computing b² - 4ac = 25 - 16 = 11 as negative, incorrectly stating the equation has no real solutions