If `sqrt(x) + (1)/(sqrt(x)) = sqrt(6)`, then `x^(2) + (1)/(x^(2))` is equal to :

Options

A. A)18
B. B)14
C. C)16
D. D)12

Answer

To solve the equation \( \sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{6} \) and find \( x^2 + \frac{1}{x^2} \), we can follow these steps:

### Step 1: Let \( y = \sqrt{x} \)
We start by substituting \( y \) for \( \sqrt{x} \). This gives us the equation:
\[ y + \frac{1}{y} = \sqrt{6} \]

### Step 2: Square both sides
Next, we square both sides of the equation to eliminate the fraction:
\[ \left(y + \frac{1}{y}\right)^2 = (\sqrt{6})^2 \]
This simplifies to:
\[ y^2 + 2 + \frac{1}{y^2} = 6 \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation to isolate \( y^2 + \frac{1}{y^2} \):
\[ y^2 + \frac{1}{y^2} = 6 - 2 \]
This simplifies to:
\[ y^2 + \frac{1}{y^2} = 4 \]

### Step 4: Substitute back to find \( x^2 + \frac{1}{x^2} \)
Since \( y = \sqrt{x} \), we have \( y^2 = x \). Therefore, we can express \( x^2 + \frac{1}{x^2} \) in terms of \( y \):
\[ x^2 + \frac{1}{x^2} = y^4 + \frac{1}{y^4} \]
To find \( y^4 + \frac{1}{y^4} \), we can use the identity:
\[ y^4 + \frac{1}{y^4} = \left(y^2 + \frac{1}{y^2}\right)^2 - 2 \]
Substituting \( y^2 + \frac{1}{y^2} = 4 \):
\[ y^4 + \frac{1}{y^4} = 4^2 - 2 = 16 - 2 = 14 \]

### Final Answer
Thus, we find that:
\[ x^2 + \frac{1}{x^2} = 14 \]