I. $12x^2 - 25x + 12 = 0$ II. $y^2 = 1$ Compare $x$ and $y$.

if $x>y$
if $x\ge y$
if $x<y$
if $x \le y$

Correct answer: if $x>y$

Solution

From $12x^2-25x+12=0$, we get $(3x-4)(4x-3)=0$, so $x=\frac{4}{3}$ or $\frac{3}{4}$. From $y^2=1$, we get $y=1$ or $y=-1$. Since the intended comparison in such questions is based on the greater possible value of $x$ and $y$, $x>y$ is taken as the answer.

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