LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
14 views
in Aptitude by (30.0k points)
closed by

In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.

I. \(6{x^2} - 25x + 14 = 0\)

II. \(9{y^2} - 9y + 2 = 0\)


1. x > y
2. x < y
3. x ≥ y
4. x ≤ y
5. x = y or relationship between x and y cannot be established

1 Answer

0 votes
by (54.3k points)
selected by
 
Best answer
Correct Answer - Option 3 : x ≥ y

I. \(6{x^2} - 25x + 14 = 0\)

\( \Rightarrow 2x\left( {3x - 2} \right) - 7\left( {3x - 2} \right) = 0\)

\( \Rightarrow \left( {2x - 7} \right)\left( {3x - 2} \right) = 0\)

\( \Rightarrow x = \frac{7}{2}\) or \(\frac{2}{3}\)

 

II. \(9{y^2} - 9y + 2 = 0\)

\( \Rightarrow 9{y^2} - 6y - 3y + 2 = 0\)

\( \Rightarrow 3y\left( {3y - 2} \right) - 1\left( {3y - 2} \right) = 0\)

\( \Rightarrow \left( {3y - 1} \right)\left( {3y - 2} \right) = 0\)

\( \Rightarrow y = \frac{1}{3}\) or \(\frac{2}{3}\)

Comparison between x and y (via Tabulation):

Value of x

Value of y

Relation

7/2

1/3

x > y

7/2

2/3

x > y

2/3

1/3

x > y

2/3

2/3

x = y


∴ \(\,x \ge y\) 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...