Multiple choice

In the following question two equations numbered I and II are given. You have to solve both the equations and give answer thereof. $(I. ) (4x^2 - 29x + 45 = 0)$ $(II. ) (3y^2 - 19y + 28 = 0)$

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

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

Equation I: 4x²-29x+45 = (4x-9)(x-5) = 0, so x = 2.25, 5. Equation II: 3y²-19y+28 = (3y-7)(y-4) = 0, so y = 7/3, 4. Compare: x=2.25 vs y=2.33 (x < y), x=2.25 vs y=4 (x < y), x=5 vs y=2.33 (x > y), x=5 vs y=4 (x > y). Since x can be less than or greater than y, relationship cannot be established. Option E correct.

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