ITPEC FE Subject B October 2024 Question 6

Quadratic equation: mapping the discriminant formula and branching logic to pseudocode.

The discriminant of ax² + bx + c = 0 is b² - 4ac.

Blank A — must compute b² - 4ac. Since b² means b × b (not b × 2), the correct expression is b × b - 4 × a × c. This eliminates options (a)–(d), which all use b × 2.

Blank B — the if branch computes two roots using sqrt(discriminant), which is only valid when discriminant > 0.

Blank C — the elseif branch computes a single repeated root −b / 2a, which applies when discriminant = 0.

Matching: A = b × b - 4 × a × c, B = discriminant > 0, C = discriminant = 0 → (h).

Why not others:
- (a)–(d) — Use b × 2 instead of b × b; this computes 2b - 4ac, not the discriminant b² - 4ac
- (e) — B = discriminant < 0 would attempt sqrt of a negative number; C = discriminant = 0 is correct but B is wrong
- (f) — B = discriminant = 0 and C = discriminant < 0 reverses the logic; the two-root branch needs > 0, not = 0
- (g) — B = discriminant = 0 puts the single-root case in the two-root branch, which is incorrect

Key rule: b² in code is b × b, not b × 2 — a common exam trap. The two-root case requires discriminant > 0; the single-root case requires discriminant = 0.

AI-generated — may contain errors