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. ` 63 x^(2) + 172 x + 117 = 0 `
II. ` 30 y^(2) + 162 y + 216 = 0 `

Options

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

Answer

To solve the given equations and find the relation between \(x\) and \(y\), we will follow these steps:

### Step 1: Solve the first equation for \(x\)

The first equation is:
\[ 63x^2 + 172x + 117 = 0 \]
We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 63\), \(b = 172\), and \(c = 117\).

### Step 2: Calculate the discriminant

First, we calculate the discriminant \(D\):
\[ D = b^2 - 4ac = 172^2 - 4 \cdot 63 \cdot 117 \]
Calculating \(172^2\):
\[ 172^2 = 29584 \]
Calculating \(4 \cdot 63 \cdot 117\):
\[ 4 \cdot 63 \cdot 117 = 29484 \]
Now, substituting back into the discriminant:
\[ D = 29584 - 29484 = 100 \]

### Step 3: Substitute values into the quadratic formula

Now we can substitute \(D\) back into the quadratic formula:
\[ x = \frac{-172 \pm \sqrt{100}}{2 \cdot 63} \]
Calculating \(\sqrt{100}\):
\[ \sqrt{100} = 10 \]
Now substituting:
\[ x = \frac{-172 \pm 10}{126} \]
This gives us two values for \(x\):
1. \(x_1 = \frac{-172 + 10}{126} = \frac{-162}{126} = -1.2857\)
2. \(x_2 = \frac{-172 - 10}{126} = \frac{-182}{126} = -1.4444\)

### Step 4: Solve the second equation for \(y\)

The second equation is:
\[ 30y^2 + 162y + 216 = 0 \]
Using the quadratic formula again:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 30\), \(b = 162\), and \(c = 216\).

### Step 5: Calculate the discriminant for \(y\)

Calculating the discriminant \(D\):
\[ D = 162^2 - 4 \cdot 30 \cdot 216 \]
Calculating \(162^2\):
\[ 162^2 = 26244 \]
Calculating \(4 \cdot 30 \cdot 216\):
\[ 4 \cdot 30 \cdot 216 = 25920 \]
Now substituting back into the discriminant:
\[ D = 26244 - 25920 = 324 \]

### Step 6: Substitute values into the quadratic formula for \(y\)

Now we can substitute \(D\) back into the quadratic formula:
\[ y = \frac{-162 \pm \sqrt{324}}{2 \cdot 30} \]
Calculating \(\sqrt{324}\):
\[ \sqrt{324} = 18 \]
Now substituting:
\[ y = \frac{-162 \pm 18}{60} \]
This gives us two values for \(y\):
1. \(y_1 = \frac{-162 + 18}{60} = \frac{-144}{60} = -2.4\)
2. \(y_2 = \frac{-162 - 18}{60} = \frac{-180}{60} = -3\)

### Step 7: Compare the values of \(x\) and \(y\)

Now we have:
- \(x_1 = -1.2857\), \(x_2 = -1.4444\)
- \(y_1 = -2.4\), \(y_2 = -3\)

Both values of \(x\) are greater than both values of \(y\):
- \(x_1 > y_1\)
- \(x_1 > y_2\)
- \(x_2 > y_1\)
- \(x_2 > y_2\)

### Conclusion

Thus, we conclude that:
\[ \text{Relation: } x > y \]