1.1 Quadratics

Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 1: Pure Mathematics 1 · official syllabus

QUADRATICS LAB

Learn it by playing it

Four short missions on a live curve. No reading first - grab a slider, watch what the graph does, and figure out the rule. When a mission glows green, read why it worked, then take the next one.

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Mission 1 of 4

Bullseye

Land the vertex on the glowing ring at (3, 1). You only get the h and k sliders - the ring forgives one notch either way.

Vertex now at (1, -2) - target (3, 1)

Study notes

Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.

Completing the square 1.1

  • Any quadratic ax² + bx + c can be written in completed-square form a(x − h)² + k.
  • Quick route when a = 1: halve the coefficient of x. x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.
  • When a is not 1, factor it out of the x terms first: 2x² + 8x + 1 = 2(x² + 4x) + 1 = 2(x + 2)² − 8 + 1 = 2(x + 2)² − 7.
  • The completed-square form gives the vertex directly: y = a(x − h)² + k has its turning point at (h, k), and the line of symmetry is x = h.
  • Use the vertex to sketch the graph and to read off the maximum or minimum value of the function.

The discriminant 1.1

  • For a quadratic ax² + bx + c = 0 (so a ≠ 0), the discriminant is b² − 4ac.
  • b² − 4ac > 0: two distinct real roots. b² − 4ac = 0: one repeated root. b² − 4ac < 0: no real roots.
  • The same test tells you how many times a line meets a curve: substitute, then examine the discriminant of the resulting quadratic.
  • Questions often reverse this: given a repeated root, set b² − 4ac = 0 and solve for the unknown coefficient.

Solving quadratic equations 1.1

  • Try factorising first: find two numbers that multiply to give c and add to give b when a = 1.
  • The formula x = (−b ± √(b² − 4ac)) / 2a always works, including when roots are not rational.
  • Completing the square also solves any quadratic and is the method behind the formula.
  • A quadratic equation can have two, one (repeated) or no real solutions - match the method to what the question asks for (exact surd answers usually want the formula or completing the square).

Quadratic inequalities 1.1

  • Solve the equation first to find the critical values, then decide which intervals satisfy the inequality.
  • Sketch the parabola: for a positive x² coefficient, ax² + bx + c < 0 between the roots and > 0 outside them.
  • Write the answer as an inequality (for example −1 < x < 5) or in interval form, and keep strict vs inclusive signs consistent with the question.

Simultaneous equations: one linear, one quadratic 1.1

  • Rearrange the linear equation for one variable and substitute into the quadratic.
  • This gives a quadratic in one variable; solve it, then back-substitute each value to find the matching partner value.
  • Solutions come in pairs of (x, y) values - give both pairs, and check them in the original equations.
  • Geometrically these are the intersection points of a line and a curve, so there can be two, one (tangent) or no solutions.

Equations that are quadratic in a function of x 1.1

  • Spot a hidden quadratic when the powers are doubled: x⁴ − 5x² + 4 = 0 is quadratic in x².
  • Substitute u = the inner expression (u = x², or u = tan x), solve the quadratic in u, then solve for x from each value of u.
  • Expect up to four solutions from an x⁴ equation: u = 1 and u = 4 give x = ±1 and x = ±2.
  • Reject any value of u that the inner expression cannot take (for example u = x² cannot be negative).

EduTribe study notes - original material aligned to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026-2027). Automated checks completed; not yet reviewed by an educator.

Free explore - your own experiments

Explore: play with the parabola

Move the sliders and watch y = a(x − h)² + k change. The green dot is the vertex; the dashed green line is the axis of symmetry; pink dots are the roots. Then add the orange line and see when it meets the curve - that is the discriminant working.

Expanded: y = 1x² − 2x − 1 · Vertex (1, -2) · Axis x = 1

Discriminant b² − 4ac = 8 → two distinct real roots

About the 3D view: your parabola is a flat 2D curve. As a visual aid we build an extra 3D surface by rotating the curve around its axis of symmetry, giving z = a((x − h)² + y²) + k - a bowl when a > 0, a dome when a < 0. The highlighted slice of that surface (y = 0) is exactly your parabola. The surface is only a picture - the maths is the 2D curve. Drag to spin; nothing moves on its own.

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