Question

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

Answer 1

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

∴ \(\,x \ge y\)