I. $3x^2 + 20x + 25 = 0$ II. $3y^2 + 14y + 8 = 0$

if x > y
if x ≥ y
if x < y
if x ≤ y

Correct answer: if x < y

Solution

Equation I factors as $(3x+5)(x+5)=0$, so $x=-\tfrac53$ or $x=-5$. Equation II factors as $(3y+2)(y+4)=0$, so $y=-\tfrac23$ or $y=-4$. Since every possible value of $x$ is less than every possible value of $y$, the relation is $x<y$.

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