If ` x + y = 4, x^(2) + y^(2) = 14 and x gt y` then the correct value of x and y is

Options

A. `2 + sqrt(3), 2 - sqrt(3)`
B. ` 2 - sqrt(2) , sqrt(3)`
C. 3,1
D. `2 + sqrt(3) , 2 sqrt(2)`

Answer

To solve the problem step by step, we start with the given equations:

1. \( x + y = 4 \) (Equation 1)
2. \( x^2 + y^2 = 14 \) (Equation 2)

### Step 1: Square Equation 1
First, we square Equation 1:

\[ (x + y)^2 = 4^2 \]

This expands to:

\[ x^2 + 2xy + y^2 = 16 \]

### Step 2: Substitute Equation 2 into the Expanded Equation
Now, we can substitute the value of \( x^2 + y^2 \) from Equation 2 into the expanded equation:

\[ 14 + 2xy = 16 \]

### Step 3: Solve for \( xy \)
Now, we can isolate \( 2xy \):

\[ 2xy = 16 - 14 \]
\[ 2xy = 2 \]
\[ xy = 1 \]

### Step 4: Set Up a Quadratic Equation
Now we have two equations:
1. \( x + y = 4 \)
2. \( xy = 1 \)

We can express \( y \) in terms of \( x \) using Equation 1:

\[ y = 4 - x \]

Substituting this into the product equation \( xy = 1 \):

\[ x(4 - x) = 1 \]

This simplifies to:

\[ 4x - x^2 = 1 \]
\[ x^2 - 4x + 1 = 0 \]

### Step 5: Solve the Quadratic Equation
Now we can solve the quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -4, c = 1 \):

\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{4 \pm \sqrt{16 - 4}}{2} \]
\[ x = \frac{4 \pm \sqrt{12}}{2} \]
\[ x = \frac{4 \pm 2\sqrt{3}}{2} \]
\[ x = 2 \pm \sqrt{3} \]

### Step 6: Determine Values of \( x \) and \( y \)
Thus, the two possible values for \( x \) are:

1. \( x = 2 + \sqrt{3} \)
2. \( x = 2 - \sqrt{3} \)

Since we are given that \( x > y \), we take:

\[ x = 2 + \sqrt{3} \]

Now, substituting back to find \( y \):

\[ y = 4 - x = 4 - (2 + \sqrt{3}) = 2 - \sqrt{3} \]

### Final Values
Thus, the correct values of \( x \) and \( y \) are:

\[ x = 2 + \sqrt{3}, \quad y = 2 - \sqrt{3} \]