Solve the given quadratic equations and mark the correct option. I. $(3x - 2)^2 = 16$ II. $y^2 + 7y + 12 = 0$

x < y
x = y or no relation can be established between x and y.
x > y
x \le y

Correct answer: x > y

Solution

From $(3x-2)^2=16$, we get $3x-2=\pm 4$, so $x=2$ or $x=-\frac{2}{3}$. From $y^2+7y+12=0$, we get $(y+3)(y+4)=0$, so $y=-3$ or $y=-4$. In all possible cases, x is greater than y.

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