I. $2x^2 + 17x + 35 = 0 \Rightarrow (2x + 7)(x + 5) = 0 \Rightarrow x = -\frac{7}{2}, -5$ II. $3y^2 + 17y + 24 = 0 \Rightarrow (y + 3)(3y + 8) = 0 \Rightarrow y = -3, -\frac{8}{3}$ Compare $x$ and $y$.

x < y
x > y
x ≤ y
x ≥ y

Correct answer: x < y

Solution

The roots of x are -7/2 and -5, while the roots of y are -3 and -8/3. Since both possible values of x are less than both possible values of y, we conclude that x < y.

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