I. $x^2 - 16x + 39 = 0$ II. $y^2 - 33y + 270 = 0$ A) If $x > y$ B) If $x \ge y$ C) If $x < y$ D) If $x \le y$

If x > y
If x ≥ y
If x < y
If x ≤ y

Correct answer: If x < y

Solution

Solving $x^2-16x+39=0$ gives roots 13 and 3. Solving $y^2-33y+270=0$ gives roots 18 and 15. In every valid pairing, $x$ is less than $y$, so the correct relation is $x<y$.

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