In the following questions two equations numbered I and II are given. You have to solve both the equations and ————— (I. \space 20x^{2} - x - 12 = 0 ) (II.\space 20y^{2} + 27y + 9 = 0)

Equation I: 20x² - x - 12 = 0. Using quadratic formula: x = [1 ± √(1 + 960)] ÷ 40 = [1 ± √961] ÷ 40 = [1 ± 31] ÷ 40. So x = 32/40 = 4/5 = 0.8, or x = -30/40 = -3/4 = -0.75. Equation II: 20y² + 27y + 9 = 0. Using quadratic formula: y = [-27 ± √(729 - 720)] ÷ 40 = [-27 ± √9] ÷ 40 = [-27 ± 3] ÷ 40. So y = -24/40 = -3/5 = -0.6, or y = -30/40 = -3/4 = -0.75. Comparing: x values are 0.8 and -0.75. y values are -0.6 and -0.75. Since x can be 0.8 > -0.6 or x can be -0.75 = -0.75, the relationship cannot be uniquely established. Option E is correct.