If `sqrt(x) + (1)/(sqrt(x)) = 2sqrt(2)`, then `x^(2) + (1)/(x^(2))` is equal to :

Options

A. 34
B. 64
C. 36
D. 32

Answer

To solve the equation \( \sqrt{x} + \frac{1}{\sqrt{x}} = 2\sqrt{2} \) and find \( x^2 + \frac{1}{x^2} \), we can follow these steps:

### Step 1: Introduce a substitution
Let \( y = \sqrt{x} \). Then, the equation becomes:
\[ y + \frac{1}{y} = 2\sqrt{2} \]

### Step 2: Square both sides
Now, we square both sides of the equation:
\[ \left(y + \frac{1}{y}\right)^2 = (2\sqrt{2})^2 \]
This simplifies to:
\[ y^2 + 2 + \frac{1}{y^2} = 8 \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation to isolate \( y^2 + \frac{1}{y^2} \):
\[ y^2 + \frac{1}{y^2} = 8 - 2 \]
\[ y^2 + \frac{1}{y^2} = 6 \]

### Step 4: Relate back to \( x \)
Since \( y = \sqrt{x} \), we have:
\[ y^2 = x \quad \text{and} \quad \frac{1}{y^2} = \frac{1}{x} \]
Thus:
\[ x^2 + \frac{1}{x^2} = (y^2)^2 + \left(\frac{1}{y^2}\right)^2 \]

### Step 5: Use the identity
We can use the identity:
\[ a^2 + b^2 = (a + b)^2 - 2ab \]
where \( a = y^2 \) and \( b = \frac{1}{y^2} \). We already know \( y^2 + \frac{1}{y^2} = 6 \) and \( ab = 1 \):
\[ x^2 + \frac{1}{x^2} = (y^2 + \frac{1}{y^2})^2 - 2 \]
Substituting the known values:
\[ x^2 + \frac{1}{x^2} = 6^2 - 2 = 36 - 2 = 34 \]

### Final Result
Thus, the value of \( x^2 + \frac{1}{x^2} \) is:
\[ \boxed{34} \]

---