Multiple choice

Find the value of variables and state the correct relationship- (6x^2 + x - 1 = 0) (6y^2 + 7y - 3 = 0)

  1. $\(x < y\)$
  2. $\(x > y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

  5. x = y or Relation cannot be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Solve 6x^2 + x - 1 = 0: x = [-1 ± √(1 + 24)]/12 = (-1 ± 5)/12, so x = 1/3 or x = -1/2. Solve 6y^2 + 7y - 3 = 0: y = [-7 ± √(49 + 72)]/12 = (-7 ± 11)/12, so y = 1/3 or y = -3/2. Since x = 1/3 matches y = 1/3, but x = -1/2 could be greater or less than y depending on which y value is chosen, multiple relationships are possible. When both roots are 1/3, x = y. When x = -1/2 and y = 1/3, x < y. When x = -1/2 and y = -3/2, x > y. Therefore, no single relationship always holds.

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