Study guide · Priority 1

Quadratics

Discriminant · completing the square · Vieta’s formulas · vertex form · building from a table

Fixes these misses: M1 Q14 · M2 Q5, Q13, Q15, Q16, Q19, Q20, Q22 — almost half of all Math errors.

Core concepts

The three forms of a quadratic

Standard: y = ax² + bx + c — c is the y-intercept.
Factored: y = a(x − r₁)(x − r₂) — r₁, r₂ are the x-intercepts (roots).
Vertex: y = a(x − h)² + k — (h, k) is the vertex.

The discriminant — how many real solutions?

D = b² − 4ac

D > 0 → two distinct real solutions (crosses x-axis twice).
D = 0 → exactly one real solution (touches x-axis — the vertex sits on it).
D < 0 → no real solutions (never touches the x-axis).
Most "no real solutions" and "exactly one solution" questions are just discriminant conditions in disguise.

Completing the square

To turn x² + bx into a perfect square, add (b/2)².
x² + bx + (b/2)² = (x + b/2)².
This is the key move for circle equations: x² + x + y² + y = k becomes (x + ½)² + (y + ½)² = k + ½, and r = √(right side).

Vieta’s formulas (sum & product of roots)

For ax² + bx + c = 0: sum = −b/a, product = c/a

You can read the sum and product of the roots straight off the coefficients — no need to actually solve.
Great for "product of the solutions is k·(something)" questions.

Worked example — "exactly one solution"

Problem. Find a so that y = −1.5 and y = x² + 8x + a meet at exactly one point.

Set equal: x² + 8x + a = −1.5, i.e. x² + 8x + (a + 1.5) = 0.

"Exactly one solution" → discriminant = 0: 8² − 4(1)(a + 1.5) = 0.

64 − 4(a + 1.5) = 0 → a + 1.5 = 16 → a = 14.5.

So a = 14.5 (this is the actual M2 Q16 answer).

Watch out: On M2 Q20 the equation is x²+x+y²+y = 199/2 (not "1992"). After completing the square on both x and y you add ¼ twice: 199/2 + ¼ + ¼ = 100, so r = √100 = 10. Don’t skip the "+¼ on both sides" step.