Multiple choice

In each of the following questions, two equations are given. You have to solve them and $(I. \quad x^2 + 13x + 40 = 0\II. \quad y^2 + 7y + 12 = 0)$

  1. If x>y

  2. If x≥y

  3. If x<y

  4. If x≤y

  5. If x=y or relationship can not be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

I: x² + 13x + 40 = 0 gives (x + 8)(x + 5) = 0, so x = -8 or -5. II: y² + 7y + 12 = 0 gives (y + 4)(y + 3) = 0, so y = -4 or -3. Comparing: when x = -8, x < y (both -4 and -3). When x = -5, x < y (both -4 and -3). Therefore x < y.

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