Question

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

I. \(6{x^2} + 25x + 24 = 0\)

II. \(12{y^2} + 13y + 3 = 0\)

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 2 : x < y

I. \(6{x^2} + 25x + 24 = 0\)

⇒ \(6{x^2} + 16x + 9x + 24 = 0\)

⇒ \(2x\left( {3x + 8} \right) + 3\left( {3x + 8} \right) = 0\)

⇒ \(x = \frac{{ - 3}}{2}\) or \(\frac{{ - 8}}{3}\)

II. \(12{y^2} + 13y + 3 = 0\)

⇒ \(12{y^2} + 4y + 9y + 3 = 0\)

⇒ \(4y\left( {3y + 1} \right) + 3\left( {3y + 1} \right) = 0\)

⇒ \(\left( {4y + 3} \right)\left( {3y + 1} \right) = 0\)

⇒ \(y = \frac{{ - 3}}{4}\) or \( - \frac{1}{3}\)

Comparison between x and y (via Tabulation):

Value of x	Value of y	Relation
-3/2	-3/4	x < y
-3/2	-1/3	x < y
-8/3	-3/4	x < y
-8/3	-1/3	x < y

∴ Clearly x < y