Question

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

I. \(15{x^2} + 8x + 1 = 0\)

II. \(3{y^2} + 14y + 8 = 0\)

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

Answer 1

Correct Answer - Option 1 : x > y

I. \(15{x^2} + 8x + 1 = 0\)

\( \Rightarrow 15{x^2} + 5x + 3x + 1 = 0\)

\( \Rightarrow 5x\left( {3x + 1} \right) + 1\left( {3x + 1} \right) = 0\)

\( \Rightarrow \left( {3x + 1} \right)\left( {5x + 1} \right) = 0\)

\( \Rightarrow x = - \frac{1}{3}\) or \( - \frac{1}{5}\)

II. \(3{y^2} + 14y + 8 = 0\)

\( \Rightarrow 3{y^2} + 2y + 12y + 8 = 0\)

\( \Rightarrow y\left( {3y + 2} \right) + 4\left( {3y + 2} \right) = 0\)

\( \Rightarrow \left( {y + 4} \right)\left( {3y + 2} \right) = 0\)

\( \Rightarrow y = - 4\) or \( - \frac{2}{3}\)

Comparison between x and y (via Tabulation):

Value of x	Value of y	Relation
-1/3	-4	x > y
-1/3	-2/3	x > y
-1/5	-4	x > y
-1/5	-2/3	x > y

∴ Clearly y < x