ITPEC FE Subject B April 2026 Question 6

A is b and B is sqrt(-val).

The program works with the normalised form x² − 2bx + c = 0, where val holds b × b − c:

• •Blank A — Case 1 applies when val is 0, and the stated double root is simply b
• •Blank B — Case 3 applies when val is negative, and the roots are b ± i√(c − b²). Since val is b² − c, the quantity c − b² equals -val, which is positive here, so sqrt(-val) is the value tmp must hold

Answer (d)

Why not others:
- (a) and (b) — use -b for the double root; the minus sign was already absorbed when b was defined as -b1 ÷ (2 × a1), so negating again flips the root
- (a) and (c) — pass val to sqrt in the imaginary branch, but val is negative there and sqrt is only defined for non-negative arguments; putting a minus outside the call does not fix the argument inside it

Key rule: Track which quantity is negative. In the imaginary case the negation belongs inside the square root, turning b² − c back into c − b².

AI-generated — may contain errors