$I . x^{2} + 5 x + 6 = 0$
$I I . y^{2} + 3 y + 2 = 0$

x > y

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\geq

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x < y

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\leq

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x = y or the relationship cannot be established

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Solution

The correct option is D x $\leq$ y
Equation I
$x^{2} + 5 x + 6 = 0$
$\Rightarrow x^{2} + 2 x + 3 x + 6 = 0$
$\Rightarrow x (x + 2) + 3 (x + 2) = 0$
$\Rightarrow (x + 3) (x + 2) = 0$
$∴ x = - 3 o r - 2$
Equation II
$y^{2} + 3 y + 2 = 0$
$\Rightarrow y^{2} + y + y + 2 = 0$
$\Rightarrow y (y + 2) + 1 (y + 2) = 0$
$\Rightarrow (y + 1) (y + 2) = 0$
$∴ y = - 1 o r - 2$
Hence, x $\leq$ y

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