Multiple choice

In each of the following questions, two equations are given. You have to solve them and $( (I) \ 3x^2 – 4x – 4 = 0 \ \ \ \ \ \ \ \ \ \ (II)\ 3y^2 + 14y + 8 = 0 )$

  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
B Correct answer
Explanation

Solve (I): 3x² - 4x - 4 = 0 factors to (3x + 2)(x - 2) = 0, giving x = 2 or x = -2/3. Solve (II): 3y² + 14y + 8 = 0 factors to (3y + 2)(y + 4) = 0, giving y = -2/3 or y = -4. Comparing: x = 2 > both y values (x ≥ y true); x = -2/3 = -2/3 (x ≥ y true); x = -2/3 > -4 (x ≥ y true). So x ≥ y always.

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