I. $x^2 + x - 240 = 0$ II. $y^2 = 256$ A) $x \ge y$ B) $x \le y$ C) if $x = y$ or no relation can be established between $x$ and $y$. D) $x < y$

$x \ge y$
$x \le y$
if $x = y$ or no relation can be established between $x$ and $y$.
$x < y$

Correct answer: if $x = y$ or no relation can be established between $x$ and $y$.

Solution

From $x^2+x-240=0$, we get $(x+16)(x-15)=0$, so $x=15$ or $x=-16$. From $y^2=256$, we get $y=16$ or $y=-16$. Since the possible values overlap and vary, a definite relation cannot always be established; in one case $x=y=-16$.

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