Directions: In each question two equations numbered I and II are given. You have to solve both the equations and mark the answer
I. 6x2 – 23√3x + 60 = 0
II. 2y2 + 3 √3y – 15 = 0
if x ≥ y
Incorrect Answerif x < y
Incorrect Answerif x ≤ y
Incorrect Answerif x = y or the relation between x and y can't be determined
Incorrect AnswerExplanation:
I. 6x2 – 23√3x + 60 = 0
6x2 – 8√3x – 15√3x + 60 = 0
2x(3x – 4√3) – 5√3 (3x – 4√3) = 0
(2x – 5√3)(3x – 4√3) = 0
x = 2.5√3,1.33√3
II. 2y2 + 3√3y – 15 = 0
2y2 + 5√3y – 2√3y – 15 = 0
y (2y + 5√3) – √3 (2y + 5√3) = 0
(2y + 5√3)(y – √3) = 0
y = √3,– 2.5√3
Therefore, for any value of x and any value of y
x > y
By: Munesh Kumari ProfileResourcesReport error