In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer
I. ` 6 x^(2) + 5 x + 1 = 0 `
II. ` 20 y ^(2) + 9 y + 1 = 0 `

Options

A. If x= y or no relation
B. If ` x gt y `
C. If ` x lt y `
D. If `x ge y`

Answer

To solve the given equations, we will follow these steps:

### Step 1: Solve Equation I
The first equation is:
\[ 6x^2 + 5x + 1 = 0 \]

This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 6 \), \( b = 5 \), and \( c = 1 \).

#### Step 1.1: Find the Discriminant
The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = 5^2 - 4 \cdot 6 \cdot 1 = 25 - 24 = 1 \]

#### Step 1.2: Use the Quadratic Formula
The roots of the equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ x = \frac{-5 \pm \sqrt{1}}{2 \cdot 6} \]
\[ x = \frac{-5 \pm 1}{12} \]

Calculating the two possible values for \( x \):
1. \( x_1 = \frac{-5 + 1}{12} = \frac{-4}{12} = -\frac{1}{3} \)
2. \( x_2 = \frac{-5 - 1}{12} = \frac{-6}{12} = -\frac{1}{2} \)

### Step 2: Solve Equation II
The second equation is:
\[ 20y^2 + 9y + 1 = 0 \]

This is also a quadratic equation where \( a = 20 \), \( b = 9 \), and \( c = 1 \).

#### Step 2.1: Find the Discriminant
Calculating the discriminant:
\[ D = b^2 - 4ac \]
Substituting the values:
\[ D = 9^2 - 4 \cdot 20 \cdot 1 = 81 - 80 = 1 \]

#### Step 2.2: Use the Quadratic Formula
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{D}}{2a} \]
Substituting the values:
\[ y = \frac{-9 \pm \sqrt{1}}{2 \cdot 20} \]
\[ y = \frac{-9 \pm 1}{40} \]

Calculating the two possible values for \( y \):
1. \( y_1 = \frac{-9 + 1}{40} = \frac{-8}{40} = -\frac{1}{5} \)
2. \( y_2 = \frac{-9 - 1}{40} = \frac{-10}{40} = -\frac{1}{4} \)

### Step 3: Compare the Values of x and y
We have the following values:
- From Equation I: \( x_1 = -\frac{1}{3} \) and \( x_2 = -\frac{1}{2} \)
- From Equation II: \( y_1 = -\frac{1}{5} \) and \( y_2 = -\frac{1}{4} \)

Now, we compare:
1. For \( x_1 = -\frac{1}{3} \) and \( y_1 = -\frac{1}{5} \):
\[ -\frac{1}{3} < -\frac{1}{5} \] (True)

2. For \( x_2 = -\frac{1}{2} \) and \( y_2 = -\frac{1}{4} \):
\[ -\frac{1}{2} < -\frac{1}{4} \] (True)

### Conclusion
In both cases, \( x < y \). Therefore, the final conclusion is:
\[ x < y \]