Multiple choice

Directions: In the following question, two equations numbered (I) and (II) are given. Student should solve both the equations and mark appropriate answer. I. 12x2 + 15x + 3 = 0 II. 14y2 + 15y + 4 = 0

  1. If x = y or no relation can be established

  2. If x > y

  3. If x < y

  4. If x ≥ y

  5. If x ≤ y
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation I: 12x^2 + 15x + 3 = 0 -> 4x^2 + 5x + 1 = 0 -> (4x+1)(x+1) = 0. Roots are -0.25, -1. Equation II: 14y^2 + 15y + 4 = 0 -> (2y+1)(7y+4) = 0. Roots are -0.5, -0.57. Comparing roots: -0.25 > -0.5 and -0.25 > -0.57, but -1 < -0.5. No relation can be established.

AI explanation

Factoring equation I by taking 3 common gives 3(4x^2 + 5x + 1) = 0, which factors into 3(4x + 1)(x + 1) = 0, yielding x = -1/4 and x = -1. Factoring equation II gives (7y + 4)(2y + 1) = 0, yielding y = -4/7 and y = -1/2. Comparing the values, -1/4 is greater than both -4/7 and -1/2, but -1 is less than both -4/7 and -1/2, so no relation can be established.

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