Multiple choice

In each of the following questions, two equations are given. You have to solve them and give the answer. $I. (x^2 + 13x + 30 = 0) II. (y^2 – 13y – 30 = 0)$

  1. If x > y

  2. If x ≤ y

  3. If x < y

  4. If x ≥ y

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

Solving equation I: x² + 13x + 30 = (x+10)(x+3) = 0, so x = -10 or -3. Solving equation II: y² - 13y - 30 = 0. Using quadratic formula: y = (13 ± √(169 + 120))/2 = (13 ± √289)/2 = (13 ± 17)/2, so y = 15 or y = -2. Comparing all combinations: when x = -10, x < y for both y values (since -10 < 15 and -10 < -2). When x = -3, x < y for y = 15, but x > y for y = -2. Therefore, we can only definitively say that x ≤ y is always true.

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