Multiple choice

Find the value of variables and state the correct relationship-( (3x^2+20x+12) =0 )( (3y^2-y-2)=0 )

  1. x < y

  2. x > y

  3. x ≤ y

  4. x ≥ y

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

For 3x² + 20x + 12 = 0, x = -2/3 or x = -6. For 3y² - y - 2 = 0, y = 1 or y = -2/3. When x = -2/3, y = 1 gives x < y; x = -2/3, y = -2/3 gives x = y. When x = -6, y = 1 gives x < y; x = -6, y = -2/3 gives x < y. In all cases x ≤ y is true.

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