I. $2x^2 - 19x + 45 = 0$ II. $2y^2 - 9y + 4 = 0$ 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

For $2x^2 - 19x + 45 = 0$, the roots are $5$ and $\frac{9}{2}$. For $2y^2 - 9y + 4 = 0$, the roots are $4$ and $\frac{1}{2}$. In every possible pairing, $x$ is greater than $y$, so the correct relation is $x > y$.

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