If `x=sqrt(2+sqrt(2+sqrt(2+........oo)))` and `y=sqrt(2sqrt(2sqrt(2........oo)))` find `x-y`

Options

A. `2x = y`
B. `3x lt y`
C. `4x = 5y`
D. `x = y`

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) defined as follows:

1. \( x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} \)
2. \( y = \sqrt{2 \cdot \sqrt{2 \cdot \sqrt{2 \cdots}}} \)

### Step 1: Solve for \( x \)

We start with the equation for \( x \):
\[ x = \sqrt{2 + x} \]
Next, we square both sides to eliminate the square root:
\[ x^2 = 2 + x \]
Rearranging this gives us a quadratic equation:
\[ x^2 - x - 2 = 0 \]

### Step 2: Factor the quadratic equation

To solve the quadratic equation, we can factor it:
\[ (x - 2)(x + 1) = 0 \]
Setting each factor to zero gives us:
\[ x - 2 = 0 \quad \text{or} \quad x + 1 = 0 \]
Thus, we find:
\[ x = 2 \quad \text{or} \quad x = -1 \]
Since \( x \) must be positive (as it is defined as a square root), we have:
\[ x = 2 \]

### Step 3: Solve for \( y \)

Now, we consider \( y \):
\[ y = \sqrt{2 \cdot y} \]
Squaring both sides gives:
\[ y^2 = 2y \]
Rearranging this gives us:
\[ y^2 - 2y = 0 \]
Factoring out \( y \):
\[ y(y - 2) = 0 \]
Setting each factor to zero gives:
\[ y = 0 \quad \text{or} \quad y = 2 \]
Since \( y \) must also be positive, we have:
\[ y = 2 \]

### Step 4: Calculate \( x - y \)

Now we can find \( x - y \):
\[ x - y = 2 - 2 = 0 \]

### Final Answer
Thus, the value of \( x - y \) is:
\[ \boxed{0} \]