In each of these questions, two equations (I) and (II) are given . You have to solve both the equations and give answer
I. ` 6 x ^(2) - 49 x + 99 = 0 `
II.` 5 y^(2) + 17 y + 14 = 0 `

Options

A. If ` x gt y `
B. If ` x lt y `
C. If ` x ge Y`
D. If ` x le y `

Answer

To solve the given equations step by step, we will start with Equation I and then move on to Equation II.

### Step 1: Solve Equation I
The first equation is:
\[ 6x^2 - 49x + 99 = 0 \]

#### Step 1.1: Use the Quadratic Formula
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6 \), \( b = -49 \), and \( c = 99 \).

#### Step 1.2: Calculate the Discriminant
First, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (-49)^2 - 4 \cdot 6 \cdot 99 \]
\[ D = 2401 - 2376 = 25 \]

#### Step 1.3: Calculate the Roots
Now, we can find the roots using the quadratic formula:
\[ x = \frac{-(-49) \pm \sqrt{25}}{2 \cdot 6} \]
\[ x = \frac{49 \pm 5}{12} \]

Calculating the two possible values for \( x \):
1. \( x_1 = \frac{49 + 5}{12} = \frac{54}{12} = 4.5 \)
2. \( x_2 = \frac{49 - 5}{12} = \frac{44}{12} = \frac{11}{3} \approx 3.67 \)

### Step 2: Solve Equation II
The second equation is:
\[ 5y^2 + 17y + 14 = 0 \]

#### Step 2.1: Use the Quadratic Formula
Again, we use the quadratic formula where \( a = 5 \), \( b = 17 \), and \( c = 14 \).

#### Step 2.2: Calculate the Discriminant
Calculate the discriminant \( D \):
\[ D = b^2 - 4ac = (17)^2 - 4 \cdot 5 \cdot 14 \]
\[ D = 289 - 280 = 9 \]

#### Step 2.3: Calculate the Roots
Now, we can find the roots:
\[ y = \frac{-17 \pm \sqrt{9}}{2 \cdot 5} \]
\[ y = \frac{-17 \pm 3}{10} \]

Calculating the two possible values for \( y \):
1. \( y_1 = \frac{-17 + 3}{10} = \frac{-14}{10} = -1.4 \)
2. \( y_2 = \frac{-17 - 3}{10} = \frac{-20}{10} = -2 \)

### Step 3: Compare Values of x and y
Now we have the values:
- From Equation I: \( x_1 = 4.5 \) and \( x_2 \approx 3.67 \)
- From Equation II: \( y_1 = -1.4 \) and \( y_2 = -2 \)

#### Step 3.1: Compare x and y
1. For \( x_1 = 4.5 \) and \( y_1 = -1.4 \):
- \( 4.5 > -1.4 \)

2. For \( x_1 = 4.5 \) and \( y_2 = -2 \):
- \( 4.5 > -2 \)

3. For \( x_2 \approx 3.67 \) and \( y_1 = -1.4 \):
- \( 3.67 > -1.4 \)

4. For \( x_2 \approx 3.67 \) and \( y_2 = -2 \):
- \( 3.67 > -2 \)

### Conclusion
In all comparisons, \( x \) is greater than \( y \). Therefore, we conclude that:
\[ x > y \]