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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

1 Answer

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Best answer
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

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