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

I. 4x^{2} + 12x + 9 = 0

II. 2y^{2} + 3y + 1 = 0

- x < y
- x > y
- x ≤ y
- x ≥ y
- x = y or the relation between x and y can't be established.

Option 1 : x < y

**Calculation:**

I. 4x^{2} + 12x + 9 = 0

⇒ 4x^{2} + 6x + 6x + 9 = 0

⇒ 2x(2x + 3) + 3(2x + 3) = 0

⇒ (2x + 3) (2x + 3) = 0

⇒ x = -3/2, -3/2

II. 2y^{2} + 3y + 1

⇒ 2y^{2} + 2y + 1y + 1 = 0

⇒ 2y(y + 1) + 1 (y + 1) = 0

⇒ (2y + 1) (y + 1) = 0

⇒ y = -1/2 , -1

Comparison between X and Y (via tabulation):

Value of x |
Value of y |
Relation |

-3/2 | -1/2 | x < y |

-3/2 | -1 | x < y |

-3/2 | -1/2 | x < y |

-3/2 | -1 | x < y |

**∴ x < y**

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