Multiple choice

In each of the following question, two equations (I) and (II) are given. You have to solve them and answer the question.\newline\textbf{I.}\space\space4x^{2} - 12x + 5 = 0 \newline\textbf{II.} 8y^{2} + 3y = 38

  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
E Correct answer
Explanation

Solving equation I: 4x² - 12x + 5 = 0 factors to (2x-1)(2x-5)=0, giving x=0.5 or x=2.5. Solving equation II: 8y² + 3y - 38 = 0 using the quadratic formula gives y = (-3 ± √1225)/16 = (-3 ± 35)/16, so y=2 or y=-19/8=-2.375. Comparing values: when x=0.5, x-2.375); when x=2.5, x>y (2.5>2). Since the relationship varies (xy for x=2.5 vs y=2), the relationship cannot be uniquely established.

Keep practicing — related questions