If `sqrt(x+2sqrt(x+2sqrt(x+2sqrt(x+2)))).....oo=x` Then value of x?

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the equation \( \sqrt{x + 2\sqrt{x + 2\sqrt{x + 2\sqrt{x + 2}} \ldots}} = x \), we can follow these steps:

### Step 1: Set up the equation
We start with the equation given in the problem:
\[ x = \sqrt{x + 2\sqrt{x + 2\sqrt{x + 2\sqrt{x + 2}} \ldots}} \]
This implies that the expression inside the square root is equal to \( x \) itself.

### Step 2: Simplify the expression
We can rewrite the equation as:
\[ x = \sqrt{x + 2} \]
This is because the infinite nested square roots will converge to the same value \( x \).

### Step 3: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ x^2 = x + 2 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ x^2 - x - 2 = 0 \]

### Step 5: Factor the quadratic equation
Next, we can factor the quadratic equation:
\[ (x - 2)(x + 1) = 0 \]

### Step 6: Solve for \( x \)
Setting each factor to zero gives us:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 7: Determine the valid solution
Since \( x \) must be non-negative (as it is under a square root), we discard \( x = -1 \). Thus, the only valid solution is:
\[ x = 2 \]

### Final Answer
The value of \( x \) is:
\[ \boxed{2} \]