If `sqrt(x) - (1)/(sqrt(x)) = 4`, then `x^(2) + (1)/(x^(2))` is equal to :

Options

A. 192
B. 326
C. 322
D. 256

Answer

To solve the equation \( \sqrt{x} - \frac{1}{\sqrt{x}} = 4 \) and find \( x^2 + \frac{1}{x^2} \), we will follow these steps:

### Step 1: Isolate the square root
We start with the equation:
\[ \sqrt{x} - \frac{1}{\sqrt{x}} = 4 \]
To eliminate the fraction, we can multiply both sides by \( \sqrt{x} \):
\[ (\sqrt{x})^2 - 1 = 4\sqrt{x} \]
This simplifies to:
\[ x - 1 = 4\sqrt{x} \]

### Step 2: Rearrange the equation
Now, we rearrange the equation:
\[ x - 4\sqrt{x} - 1 = 0 \]

### Step 3: Substitute \( y = \sqrt{x} \)
Let \( y = \sqrt{x} \). Then \( x = y^2 \) and we can rewrite the equation:
\[ y^2 - 4y - 1 = 0 \]

### Step 4: Solve the quadratic equation
We can solve this quadratic equation using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -4, c = -1 \):
\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ y = \frac{4 \pm \sqrt{16 + 4}}{2} \]
\[ y = \frac{4 \pm \sqrt{20}}{2} \]
\[ y = \frac{4 \pm 2\sqrt{5}}{2} \]
\[ y = 2 \pm \sqrt{5} \]

### Step 5: Find \( x \)
Since \( y = \sqrt{x} \), we have:
\[ \sqrt{x} = 2 + \sqrt{5} \quad \text{(taking the positive root)} \]
Squaring both sides gives:
\[ x = (2 + \sqrt{5})^2 \]
\[ x = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5} \]

### Step 6: Find \( x^2 + \frac{1}{x^2} \)
Now we need to find \( x^2 + \frac{1}{x^2} \). First, we find \( x + \frac{1}{x} \):
\[ x + \frac{1}{x} = (2 + \sqrt{5})^2 + \frac{1}{(2 + \sqrt{5})^2} \]

Using the identity:
\[ x + \frac{1}{x} = \sqrt{x} + \frac{1}{\sqrt{x}} = 4 + 2 = 6 \]

Now we can find \( x^2 + \frac{1}{x^2} \) using the identity:
\[ x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 \]
\[ x^2 + \frac{1}{x^2} = 6^2 - 2 = 36 - 2 = 34 \]

### Final Answer
Thus, \( x^2 + \frac{1}{x^2} = 34 \).

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