Two equations (I) and (II) are given in each questions. On the basis of these equations you have to decide the relation between `x` and `y` and give answer
I. ` 81 x^(2) - 9 x - 2 = 0 `
II. ` 56 y^(2) - 13 y - 3 = 0 `

Options

A. If ` x gt y `
B. If `x ge y`
C. If ` x lt y `
D. If `x = y` or Relation cannot be established

Answer

To determine the relationship between \( x \) and \( y \) based on the given equations, we will solve each equation step by step.

### Step 1: Solve Equation I
The first equation is:
\[ 81x^2 - 9x - 2 = 0 \]
This is a quadratic equation of the form \( ax^2 + bx + c = 0 \) where:
- \( a = 81 \)
- \( b = -9 \)
- \( c = -2 \)

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 2: Substitute Values into the Quadratic Formula
Substituting \( a \), \( b \), and \( c \) into the formula:
\[ x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 81 \cdot (-2)}}{2 \cdot 81} \]
\[ x = \frac{9 \pm \sqrt{81 + 648}}{162} \]
\[ x = \frac{9 \pm \sqrt{729}}{162} \]

### Step 3: Simplify the Square Root
Calculating the square root:
\[ \sqrt{729} = 27 \]
Now substituting back:
\[ x = \frac{9 \pm 27}{162} \]

### Step 4: Calculate the Two Possible Values for \( x \)
Calculating the two values:
1. \( x_1 = \frac{9 + 27}{162} = \frac{36}{162} = \frac{2}{9} \approx 0.222 \)
2. \( x_2 = \frac{9 - 27}{162} = \frac{-18}{162} = -\frac{1}{9} \approx -0.111 \)

### Step 5: Solve Equation II
The second equation is:
\[ 56y^2 - 13y - 3 = 0 \]
This is also a quadratic equation where:
- \( a = 56 \)
- \( b = -13 \)
- \( c = -3 \)

Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 6: Substitute Values into the Quadratic Formula for \( y \)
Substituting \( a \), \( b \), and \( c \):
\[ y = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 56 \cdot (-3)}}{2 \cdot 56} \]
\[ y = \frac{13 \pm \sqrt{169 + 672}}{112} \]
\[ y = \frac{13 \pm \sqrt{841}}{112} \]

### Step 7: Simplify the Square Root
Calculating the square root:
\[ \sqrt{841} = 29 \]
Now substituting back:
\[ y = \frac{13 \pm 29}{112} \]

### Step 8: Calculate the Two Possible Values for \( y \)
Calculating the two values:
1. \( y_1 = \frac{13 + 29}{112} = \frac{42}{112} = \frac{3}{8} \approx 0.375 \)
2. \( y_2 = \frac{13 - 29}{112} = \frac{-16}{112} = -\frac{2}{14} \approx -0.143 \)

### Step 9: Compare Values of \( x \) and \( y \)
Now we have:
- \( x_1 \approx 0.222 \), \( x_2 \approx -0.111 \)
- \( y_1 \approx 0.375 \), \( y_2 \approx -0.143 \)

Comparing the values:
- For \( x_1 \) and \( y_1 \): \( 0.222 < 0.375 \) (so \( x < y \))
- For \( x_1 \) and \( y_2 \): \( 0.222 > -0.143 \) (so \( x > y \))

### Conclusion
Since we have \( x < y \) in one case and \( x > y \) in another, the relationship between \( x \) and \( y \) cannot be established.

### Final Answer
The relation between \( x \) and \( y \) cannot be established.

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