In each of the following questions, two equations numbered I and II are given. You have to solve them and $( (I) \quad 14x^2 – 5x – 1 = 0 \\ (II) \quad 2y^2 + 3y + 1 = 0 )$ Give answer

Equation I: 14x² - 5x - 1 = 0. Using quadratic formula: x = [5 ± sqrt(25 - 4(14)(-1))]/28 = [5 ± sqrt(25 + 56)]/28 = [5 ± sqrt(81)]/28 = [5 ± 9]/28. So x = 14/28 = 1/2 or x = -4/28 = -1/7. Equation II: 2y² + 3y + 1 = 0. Factors: (2y + 1)(y + 1) = 0. So y = -1/2 or y = -1. Both values of x (1/2, -1/7) are greater than both values of y (-1/2, -1). Therefore x > y.