Multiple choice

In each of the following questions, two equations are given. You have to solve them and Give answer. (I) (14x^2 -131x +102 =0) (II) (63y^2 +268y + 285 =0)

  1. If x>y

  2. If x≥y

  3. If x<y

  4. If x≤y

  5. If x=y or relationship cannot be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equation (I): 14x² - 131x + 102 = 0. Factorizing: (2x - 9)(7x - 11) = 0, giving x = 4.5 or x = 11/7 ≈ 1.57. For equation (II): 63y² + 268y + 285 = 0. Using the quadratic formula: y = [-268 ± √(268² - 4×63×285)]/(2×63) = [-268 ± √(71824 - 71820)]/126 = [-268 ± √4]/126 = [-268 ± 2]/126, giving y = -266/126 = -1.21 or y = -270/126 = -1.43. Both x values (4.5, 1.57) are greater than both y values (-1.21, -1.43). Therefore, x > y.

Keep practicing — related questions