Mathematics (9709)
Factorising: x² + 5x + 6 = 0 → (x+2)(x+3) = 0 → x = -2 or x = -3 Quadratic Formula: x = (-b ± √(b²-4ac)) / 2a Completing the Square: Write in form (x+p)² + q
b²-4ac > 0: Two distinct real roots (graph crosses x-axis twice) b²-4ac = 0: One repeated root (graph touches x-axis) b²-4ac < 0: No real roots (graph doesn't touch x-axis)
If a > 0: U-shaped (minimum point) If a < 0: ∩-shaped (maximum point) Vertex (turning point): x = -b/2a y-intercept: substitute x = 0 x-intercepts: solve ax² + bx + c = 0
Topic 1 of 1Cambridge A Levels
Quadratics & Functions
Solving quadratic equations and understanding graphs
A quadratic equation has the form ax² + bx + c = 0 where a ≠ 0.
Methods for Solving:
The Discriminant (b² - 4ac):
Quadratic Graphs (Parabolas):
Simultaneous Equations:
One linear + one quadratic: substitute the linear into the quadratic, solve the resulting quadratic.
Key Points to Remember
- 1Quadratic formula: x = (-b ± √(b²-4ac)) / 2a
- 2Discriminant b²-4ac determines number of roots
- 3Completing the square: write as (x+p)² + q to find vertex
- 4Vertex of parabola at x = -b/2a
Pakistan Example
Maximising Profit — A Pakistani Business Problem
A Karachi food stall owner sells samosas. If he charges Rs x per samosa, he sells (200-2x) samosas per day. Revenue = x(200-2x) = 200x - 2x². This is a quadratic! To find maximum revenue: vertex at x = -200/(2×(-2)) = 50. So charging Rs 50 per samosa maximises revenue at 200(50) - 2(50²) = Rs 5,000/day. This is a real optimisation problem solved with A-Level maths. The discriminant confirms two x-intercepts at x=0 and x=100 — charging more than Rs 100 means zero sales!