Multiple choice

Solve the following equations and find the values of (x) and (y): I. (16x^2 + 20x + 6 = 0) II. (10y^2 + 38y + 24 = 0)

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

  5. x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For I: 16x^2 + 20x + 6 = 0. Using quadratic formula: x = [-20 ± √(400-384)]/32 = [-20 ± 4]/32, giving x = -16/32 = -0.5 or x = -24/32 = -0.75. For II: 10y^2 + 38y + 24 = 0. y = [-38 ± √(1444-960)]/20 = [-38 ± √484]/20 = [-38 ± 22]/20, giving y = -16/20 = -0.8 or y = -60/20 = -3. Comparing: x values (-0.5, -0.75) are both greater than y values (-0.8, -3). Thus x > y.

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