Multiple choice

Find the value of variables and state the correct relationship-(12x^2+28x+15=0 )(6y^2+y-2=0)

  1. $x < y$
  2. $x > y$
  3. $x ≤ y$
  4. $x ≥ y$

  5. x = y or Relation cannot be established
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A Correct answer
Explanation

Solve 12x^2 + 28x + 15 = 0: x = [-28 ± √(784 - 720)]/24 = (-28 ± 8)/24, so x = -20/24 = -5/6 or x = -36/24 = -3/2. Solve 6y^2 + y - 2 = 0: y = [-1 ± √(1 + 48)]/12 = (-1 ± 7)/12, so y = 6/12 = 1/2 or y = -8/12 = -2/3. Compare: x = -5/6 with y = 1/2 gives x < y (-0.833 < 0.5). x = -5/6 with y = -2/3 gives x < y (-0.833 < -0.667). x = -3/2 with y = 1/2 gives x < y (-1.5 < 0.5). x = -3/2 with y = -2/3 gives x < y (-1.5 < -0.667). In all four cases, x < y.

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