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

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

(a)

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(b)

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(d)

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(e)

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Solution

The correct option is B
(b)

On solving we get x= - $\frac{2}{3}$ or -2

y= $\frac{3}{4}$ or 4

So, clearly, x $<$ y

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Similar questions

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

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

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

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

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

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

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

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

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

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