Quadratic Equations

Quadratic Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

A quadratic equation in one variable x has the standard form ax² + bx + c = 0, with real coefficients a, b, c and a ≠ 0. Its roots are the values of x that satisfy the equation, and it always has two roots (which may coincide or be non-real). The single most important formula is the quadratic formula:

x = [−b ± √(b² − 4ac)] / 2a

The expression under the square root is the discriminant, D = b² − 4ac, and its sign tells you the nature of the roots: D > 0 gives two distinct real roots, D = 0 gives two equal real roots, and D < 0 gives two non-real complex conjugate roots. The sum of the roots is α + β = −b/a and the product is αβ = c/a. In NABTEB, factorise whenever you can spot integer factors; only fall back to the formula when factorisation is messy.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Standard form and the four solution methods

Writing ax² + bx + c = 0 with a ≠ 0 is the gateway step — most slips begin before solving even starts. Four methods are acceptable in NABTEB: (1) factorisation into (px + q)(rx + s) = 0; (2) completing the square by rewriting as a(x + b/2a)² = (b² − 4ac)/4a²; (3) the quadratic formula x = [−b ± √(b² − 4ac)] / 2a; and (4) the graphical method, where the x-intercepts of y = ax² + bx + c are the roots.

Discriminant and the nature of roots

The value of D = b² − 4ac classifies the roots before you compute them. The summary table below is the one NABTEB markers test most often:

Condition on D	Nature of roots	Parabola y = ax² + bx + c
D > 0, a perfect square	Two distinct rational roots	Cuts x-axis at two points
D > 0, not a perfect square	Two distinct irrational roots	Cuts x-axis at two points
D = 0	Two equal real roots (x = −b/2a)	Touches x-axis at one point (vertex)
D < 0	Two complex conjugate roots (no real root)	Does not meet the x-axis

Vieta’s relations

When the roots are α and β, α + β = −b/a and αβ = c/a. These let you form a new equation from given roots using x² − (sum)x + (product) = 0, and they let you find unknown coefficients when one root is known.

Typical NABTEB question patterns

Expect a 2–3 mark item asking you to solve by factorisation, a 3–4 mark item requiring the quadratic formula with a clearly non-factorable trinomial, a 3-mark discriminant/nature-of-roots question, and a 4–5 mark word problem (age, area, speed, revenue) that reduces to a quadratic. Show all working; partial credit is awarded for the correct setup.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Vertex, axis of symmetry, and maximum/minimum

Completing the square rewrites y = ax² + bx + c as y = a(x − h)² + k, where the vertex is (h, k) = (−b/2a, (4ac − b²)/4a). The axis of symmetry is the vertical line x = −b/2a. When a > 0 the parabola opens upward and k is the minimum value; when a < 0 it opens downward and k is the maximum value. Optimisation questions (maximum profit, minimum cost) live here.

Forming equations and coefficient problems

Given α and β, the equation is x² − (α + β)x + αβ = 0. Given one root and the sum or product, substitute into Vieta’s relations to recover b and c without ever solving the quadratic.

Common traps to avoid

• Forgetting to enforce a ≠ 0 before declaring an equation quadratic.
• Dropping the ± sign and reporting only one root.
• Confusing α + β = −b/a with +b/a, or αβ = c/a with −c/a.
• Forgetting to convert a word problem to standard form before solving.

Worked micro-example

Solve 2x² − 5x − 3 = 0. Here a = 2, b = −5, c = −3, so D = 25 + 24 = 49. x = [5 ± 7]/4, giving x = 3 or x = −½. Check via factorisation: (2x + 1)(x − 3) = 2x² − 6x + x − 3 = 2x² − 5x − 3 ✓.

Practice prompts

1. Find k if x² − 3kx + 9 = 0 has equal roots.
2. Form the quadratic whose roots are 2 + √3 and 2 − √3.

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