Multiple choice

$I. (8x^2 + 6x = 5) II. (12y^2 - 22y + 8 = 0)$ Solve the given equations and find the values of x and y.

  1. If x < y

  2. If x > y

  3. If x ≤ y

  4. If x ≥ y

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

For I: 8x^2 + 6x - 5 = 0. Using formula: x = [-6 ± √(36+160)]/16 = [-6 ± √196]/16 = [-6 ± 14]/16, giving x = 8/16 = 0.5 or x = -20/16 = -1.25. For II: 12y^2 - 22y + 8 = 0. y = [22 ± √(484-384)]/24 = [22 ± 10]/24, giving y = 32/24 = 4/3 ≈ 1.33 or y = 12/24 = 0.5. Comparing: x values are 0.5 and -1.25; y values are 1.33 and 0.5. Since 0.5 = 0.5 and -1.25 < both y values, x ≤ y is correct.

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