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

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

if x > y

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if x < y

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if x ≤ y

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if x = y or the relationship cannot be established

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Solution

The correct option is B
if x < y

I. 3 $x^{2}$ + 8x + 4 = 0
Or 3 $x^{2}$ + 6x + 2x + 4 = 0
Or 3x (x + 2) + 2(x + 2) = 0
Or (3x + 2) (x + 2) = 0
Therefore x = $- \frac{2}{3}$ or -2
II. 4 $y^{2}$ - 19y + 12 = 0
Or 4 $y^{2}$ - 16y - 3y + 12 = 0
Or 4y (y - 4) - 3(y - 4) = 0
Or (4y - 3) (y - 4) = 0
Therefore y = $\frac{3}{4}$ or 4
So x < y is the answer.

Suggest Corrections

Similar questions

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

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

3 x^{2} + 8 x + 4 = 0

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

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

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

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

In the following question, two equations numbered I and II are given. Find the values of x and y and choose the appropriate answer.

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

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

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