Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equations and choose the correct answer. \ I. (x^2 - 14x - 32 = 0) \ II. (y^2 + 4y - 60 = 0)

  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

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

Solving equation I: x² - 14x - 32 = 0 gives (x - 16)(x + 2) = 0, so x = 16 or x = -2. Solving equation II: y² + 4y - 60 = 0 gives (y + 10)(y - 6) = 0, so y = -10 or y = 6. Comparing all values: x = 16 > y = -10, x = 16 > y = 6, x = -2 > y = -10, x = -2 < y = 6. Since x is sometimes greater than and sometimes less than y, the relationship cannot be established.

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