Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and answer the question. $( (I) \ 3x^2 - 4x - 32 = 0 \\ (II) \ 2y^2 - 17y + 36 = 0 )$

  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
D Correct answer
Explanation

Solving equation (I): 3x²-4x-32=0. Using quadratic formula: x=(4±√(16+384))/6=(4±√400)/6=(4±20)/6, giving x=24/6=4 or x=-16/6=-8/3≈-2.67. Solving equation (II): 2y²-17y+36=0 factors to (y-9)(2y-4)=0, giving y=9 or y=2. Comparing: when x=4, y=2 gives x>y; y=9 gives x2 is false, but 4<9 and -2.67<9 and -2.67<2 are all true), the correct answer is x≤y.

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