In each of these questions, two equations I. and II. are give. you have to solve both the equations and give answer
I. ` x^(2) - 4 sqrt( 3) ( sqrt(3) + 1) x + 48 sqrt(3) = 0 `
II. ` y^(2) - 2 sqrt(5) ( sqrt(5) + 2) y + 40 sqrt( 5) = 0 `

Options

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

Answer

To solve the given equations step by step, we will first tackle Equation I and then Equation II.

### Step 1: Solve Equation I
The first equation is:
\[ x^2 - 4\sqrt{3}(\sqrt{3} + 1)x + 48\sqrt{3} = 0 \]

First, we simplify the coefficient of \(x\):
\[ 4\sqrt{3}(\sqrt{3} + 1) = 4\sqrt{3} \cdot \sqrt{3} + 4\sqrt{3} \cdot 1 = 12 + 4\sqrt{3} \]
Thus, the equation becomes:
\[ x^2 - (12 + 4\sqrt{3})x + 48\sqrt{3} = 0 \]

### Step 2: Use the Quadratic Formula
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -(12 + 4\sqrt{3})\), and \(c = 48\sqrt{3}\).

Calculating \(b^2 - 4ac\):
\[ b^2 = (12 + 4\sqrt{3})^2 = 144 + 96\sqrt{3} + 48 = 192 + 96\sqrt{3} \]
\[ 4ac = 4 \cdot 1 \cdot 48\sqrt{3} = 192\sqrt{3} \]
Thus,
\[ b^2 - 4ac = (192 + 96\sqrt{3}) - 192\sqrt{3} = 192 - 96\sqrt{3} \]

### Step 3: Substitute into the Quadratic Formula
Now substituting back into the quadratic formula:
\[ x = \frac{12 + 4\sqrt{3} \pm \sqrt{192 - 96\sqrt{3}}}{2} \]

### Step 4: Solve for y in Equation II
The second equation is:
\[ y^2 - 2\sqrt{5}(\sqrt{5} + 2)y + 40\sqrt{5} = 0 \]

First, we simplify the coefficient of \(y\):
\[ 2\sqrt{5}(\sqrt{5} + 2) = 10 + 4\sqrt{5} \]
Thus, the equation becomes:
\[ y^2 - (10 + 4\sqrt{5})y + 40\sqrt{5} = 0 \]

### Step 5: Use the Quadratic Formula for y
Using the quadratic formula again:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -(10 + 4\sqrt{5})\), and \(c = 40\sqrt{5}\).

Calculating \(b^2 - 4ac\):
\[ b^2 = (10 + 4\sqrt{5})^2 = 100 + 80\sqrt{5} + 80 = 180 + 80\sqrt{5} \]
\[ 4ac = 4 \cdot 1 \cdot 40\sqrt{5} = 160\sqrt{5} \]
Thus,
\[ b^2 - 4ac = (180 + 80\sqrt{5}) - 160\sqrt{5} = 180 - 80\sqrt{5} \]

### Step 6: Substitute into the Quadratic Formula for y
Now substituting back into the quadratic formula:
\[ y = \frac{10 + 4\sqrt{5} \pm \sqrt{180 - 80\sqrt{5}}}{2} \]

### Final Step: Summary of Solutions
The solutions for \(x\) and \(y\) are:
- For Equation I: \( x = \frac{12 + 4\sqrt{3} \pm \sqrt{192 - 96\sqrt{3}}}{2} \)
- For Equation II: \( y = \frac{10 + 4\sqrt{5} \pm \sqrt{180 - 80\sqrt{5}}}{2} \)