Multiple choice

Directions: In the following question, two equations numbered I and II are given. You have to solve both the equations and then choose the correct option. I. x2 − 18x + 45 = 0 II. 3y2 + 2y = 1

  1. x > y

  2. x < y

  3. x ≥ y

  4. x ≤ y

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

Eq I: x^2 - 18x + 45 = 0 -> (x-15)(x-3) = 0, so x = 15 or 3. Eq II: 3y^2 + 2y - 1 = 0 -> (3y-1)(y+1) = 0, so y = 1/3 or -1. Comparing values, x is always greater than y.

AI explanation

For equation I, splitting the middle term of x^2 - 18x + 45 = 0 gives x^2 - 15x - 3x + 45 = 0, leading to the factors (x - 3)(x - 15) = 0 and the roots x = 3 and x = 15. For equation II, 3y^2 + 2y - 1 = 0 factors into (3y - 1)(y + 1) = 0, giving the roots y = 1/3 and y = -1. Because both roots of x (3 and 15) are greater than both roots of y (1/3 and -1), x > y.

Keep practicing — related questions