Multiple choice

In each of the following questions, two equations I and II have been given. Solve these questions and answer I. (30x^2 + 11x + 1 = 0) II. (42y^2 + 13y + 1 = 0)

  1. If x < y

  2. If x ≤ y

  3. If x = y or the relation cannot be established

  4. If x ≥ y

  5. If x > y
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: 30x² + 11x + 1 = 0 factors to (5x+1)(6x+1)=0, giving x = -1/5 ≈ -0.2 or x = -1/6 ≈ -0.167. Equation II: 42y² + 13y + 1 = 0 factors to (6y+1)(7y+1)=0, giving y = -1/6 ≈ -0.167 or y = -1/7 ≈ -0.143. Comparing the values: x's maximum is -0.167 which equals y's minimum, and x's minimum -0.2 is less than both y values. Therefore x ≤ y is always true.

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