Solve the quadratic equations and determine the relation between x and y: Equation 1: 2x² - 10x + 12 = 0 Equation 2: 4y² - 16y + 15 = 0

Solution

From equation 1: 2x² - 10x + 12 = 0 Dividing by 2: x² - 5x + 6 = 0 Factorizing: (x - 3)(x - 2) = 0 So, x = 3 or x = 2. From equation 2: 4y² - 16y + 15 = 0 Dividing by 4: y² - 4y + 3.75 = 0 Discriminant: (-4)² - 4(1)(3.75) = 16 - 15 = 1 (positive). Roots: y = (4 ± √1) / 2 = 2.5 or 1.5. Comparing x and y: • If x = 3 and y = 2.5, x > y. • If x = 2 and y = 2.5, x < y. • If x = 3 and y = 1.5, x > y. • If x = 2 and y = 1.5, x > y. Thus, x can be both greater than or less than y. Answer: E