2x $^{2}$ + 11x + 14 = 0

4y $^{2}$ + 12y + 9 = 0

(a)

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(b)

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(c)

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(d)

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(e)

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Solution

The correct option is B
(b)

On solving, we will get

x=- $\frac{7}{2}$ or -2

y=- $\frac{3}{2}$

So, clearly, x $<$ y

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Similar questions

2x $^{2}$ + 11x + 14 = 0

4y $^{2}$ + 12y + 9 = 0

2 x^{2} + 11 x + 14 = 0

4 y^{2} + 12 y + 9 = 0

2 x^{2} + 11 x + 14

\frac{1}{2} x^{2} + \sqrt{11} x + 1 = 0

x = - 14, - 8

2 x^{2} - 11 x + 14 & 3 x^{2} - x - 10

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