If `x = 3 + 2 sqrt2 ,` find the value `sqrtx - (1)/(sqrtx)`

Answer

To solve the problem where \( x = 3 + 2\sqrt{2} \) and we need to find the value of \( \sqrt{x} - \frac{1}{\sqrt{x}} \), we will follow these steps:

### Step 1: Find \( \frac{1}{x} \)
We start with the expression for \( x \):
\[ x = 3 + 2\sqrt{2} \]
To find \( \frac{1}{x} \), we rationalize the denominator:
\[ \frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} \cdot \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} \]
Now, we will calculate the denominator using the difference of squares:
\[ (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]
Thus, we have:
\[ \frac{1}{x} = 3 - 2\sqrt{2} \]

### Step 2: Find \( x + \frac{1}{x} \)
Now we can find \( x + \frac{1}{x} \):
\[ x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 6 \]

### Step 3: Use the identity to find \( \sqrt{x} - \frac{1}{\sqrt{x}} \)
We know that:
\[ \sqrt{x} - \frac{1}{\sqrt{x}} = \sqrt{x} + \frac{1}{\sqrt{x}} - 2 \]
Let \( y = \sqrt{x} + \frac{1}{\sqrt{x}} \). From the previous step, we can find \( y^2 \):
\[ y^2 = x + 2 + \frac{1}{x} \]
Substituting \( x + \frac{1}{x} = 6 \):
\[ y^2 = 6 + 2 = 8 \]
Thus, \( y = \sqrt{8} = 2\sqrt{2} \).

### Step 4: Calculate \( \sqrt{x} - \frac{1}{\sqrt{x}} \)
Now we can find:
\[ \sqrt{x} - \frac{1}{\sqrt{x}} = y - 2 = 2\sqrt{2} - 2 \]

### Step 5: Final Result
The final answer is:
\[ \sqrt{x} - \frac{1}{\sqrt{x}} = 2(\sqrt{2} - 1) \]