Multiple choice

Find the value of variables and state the correct relation. (6x^2 + 41x + 13 = 0) (2y^2 + 27y + 81 = 0)

  1. $x > y$
  2. $x \ge y$
  3. $x < y$
  4. $x \le y$
  5. $x = y or relationship cannot be established$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For x: 6x² + 41x + 13 = 0. Discriminant = 1681 - 312 = 1369 = 37². Roots: x = [-41 ± 37]/12 → x = -4/12 = -1/3 or x = -78/12 = -6.5. For y: 2y² + 27y + 81 = 0. Discriminant = 729 - 648 = 81 = 9². Roots: y = [-27 ± 9]/4 → y = -18/4 = -4.5 or y = -36/4 = -9. Comparing: When x = -1/3, x > both y values. When x = -6.5, x > -9 but x < -4.5. Since the relationship varies, E (cannot be established) is correct.

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