Multiple choice

$I. (18x^2 + 18x + 4 = 0)$ $II. (12y^2 + 29y + 14 = 0)$ 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
D Correct answer
Explanation

For I: 18x^2 + 18x + 4 = 0. Discriminant = 324 - 288 = 36. x = [-18 ± 6]/36, giving x = -12/36 = -1/3 or x = -24/36 = -2/3. For II: 12y^2 + 29y + 14 = 0. Discriminant = 841 - 672 = 169. y = [-29 ± 13]/24, giving y = -16/24 = -2/3 or y = -42/24 = -7/4 = -1.75. Since x = -1/3, -2/3 and y = -2/3, -1.75, we have x ≥ y (because -2/3 = -2/3 and -1/3 > -2/3 > -1.75).

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