Multiple choice

In the following questions two equations I and II are given. You have to solve both the equations and give answer.(I) \ 23x^2 – 29x – 88 = 0 (II) \ 29y^2 + 33y – 48 = 0

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Equation I: 23x²-29x-88=0. Using quadratic formula: x = (29 ± √(841+8084))/46 = (29 ± √8925)/46. √8925 ≈ 94.47, so x ≈ (29 ± 94.47)/46, giving x ≈ 2.68 or x ≈ -1.43. Equation II: 29y²+33y-48=0 gives y = (-33 ± √(1089+5568))/58 = (-33 ± √6657)/58. √6657 ≈ 81.59, so y ≈ 0.84 or y ≈ -1.97. x=2.68 > y=0.84 and > y=-1.97, but x=-1.43 is between y values. Relationship can't be uniquely determined. Option E is correct.

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