Single correct option
Real numbers a, b and c satisfy a + b + c = 6 and ab + bc + ca = 9. What is the maximum possible value of c?
- A. 3
- B. 4
- C. 6
- D. 2
Show answer and solution
Answer: (B) 4
From the equations: a + b = 6 − c and ab = 9 − c(a + b) = 9 − c(6 − c) = c² − 6c + 9.
So a and b are roots of t² − (6 − c)t + (c² − 6c + 9) = 0. For a and b to be real, the discriminant must be ≥ 0:
(6 − c)² − 4(c² − 6c + 9) ≥ 0
36 − 12c + c² − 4c² + 24c − 36 ≥ 0
12c − 3c² ≥ 0 ⇒ 3c(4 − c) ≥ 0 ⇒ 0 ≤ c ≤ 4.
Maximum c = 4, attained at a = b = 1 (check: 1 + 1 + 4 = 6 and 1 + 4 + 4 = 9 ✓).
Tip: To bound one variable, treat the others as roots of a quadratic and impose discriminant ≥ 0.
So a and b are roots of t² − (6 − c)t + (c² − 6c + 9) = 0. For a and b to be real, the discriminant must be ≥ 0:
(6 − c)² − 4(c² − 6c + 9) ≥ 0
36 − 12c + c² − 4c² + 24c − 36 ≥ 0
12c − 3c² ≥ 0 ⇒ 3c(4 − c) ≥ 0 ⇒ 0 ≤ c ≤ 4.
Maximum c = 4, attained at a = b = 1 (check: 1 + 1 + 4 = 6 and 1 + 4 + 4 = 9 ✓).
Tip: To bound one variable, treat the others as roots of a quadratic and impose discriminant ≥ 0.
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