In the following question, two equations numbered I and II are given. You have to solve both the equations and choose the apropriate answer.

I. 3x $^{2}$ + 8x + 4 = 0

II. 4y $^{2}$ - 19y + 12 = 0

y > x

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x ≤ y

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y ≤ x

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x = y or the relationship cannot be established

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Solution

The correct option is A
y > x

$\begin{matrix} I . & 3 x^{2} + 8 x + 4 = 0 & \Rightarrow 3 x^{2} + 6 x + 2 x + 4 = 0 & \Rightarrow 3 x (x + 2) + 2 (x + 2) = 0 & \Rightarrow (x + 2) (3 x + 2) = 0 & ∴ x = - 2 o r - \frac{2}{3} I I . & 4 y^{2} - 19 y + 12 = 0 & \Rightarrow 4 y^{2} - 16 y - 3 y + 12 = 0 & \Rightarrow 4 y (y - 4) - 3 (y - 4) = 0 & \Rightarrow (y - 4) (4 y - 3) = 0 & ∴ y = 4 o r \frac{3}{4} C l e a r l y, x < y \end{matrix}$

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