# In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer. I. $3{x^2} + 8x + 4 = 0$

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In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. $3{x^2} + 8x + 4 = 0$

II. $4{y^2} - 19y + 12 = 0$

1. x > y
2. x < y
3. x ≥ y
4. x ≤ y
5. x = y or relationship between x and y cannot be established

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Correct Answer - Option 2 : x < y

I. $3{x^2} + 8x + 4 = 0$

$\Rightarrow 3{x^2} + 6x + 2x + 4 = 0$

$\Rightarrow 3x\left( {x + 2} \right) + 2\left( {x + 2} \right) = 0$

$\Rightarrow \left( {x + 2} \right)\left( {3x + 2} \right) = 0$

∴ x = -2 or -2/3

II. $4{y^2} - 19y + 12 = 0$

$\Rightarrow 4{y^2} - 16y - 3y + 12 = 0$

$\Rightarrow 4y\left( {y - 4} \right) - 3\left( {y - 4} \right) = 0$

$\Rightarrow \left( {y - 4} \right)\left( {4y - 3} \right) = 0$

∴ y = 4 or 3/4

Comparison between x and y (via Tabulation):

 Value of x Value of y Relation -2 4 x < y -2 3/4 x < y -2/3 4 x < y -2/3 3/4 x < y

∴ Clearly x < y