Multiple choice

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer: $(I. x^2 + 13x + 40 = 0 )$ $(II. y^2 + 7y + 10 = 0)$

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

  5. x = y or relation can’t be established
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation I: x^2 + 13x + 40 = 0 → (x+8)(x+5)=0 → x = -8, -5. Equation II: y^2 + 7y + 10 = 0 → (y+5)(y+2)=0 → y = -5, -2. Comparing: x=-8 < y=-5,-2; x=-5 = y=-5, but x=-5 < y=-2. Since x is sometimes equal but never greater than y, x <= y is correct.

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