If `x^4 + x^(-4) = 194 , x gt 0,` then the value of `x + (1)/(x)` is :

Options

A. 4
B. 6
C. 8
D. 14

Answer

To solve the equation \( x^4 + x^{-4} = 194 \) for \( x + \frac{1}{x} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^4 + x^{-4} = 194 \]
We can rewrite \( x^{-4} \) as \( \frac{1}{x^4} \):
\[ x^4 + \frac{1}{x^4} = 194 \]

### Step 2: Introduce a substitution
Let \( y = x + \frac{1}{x} \). We need to express \( x^4 + \frac{1}{x^4} \) in terms of \( y \).

### Step 3: Use the identity
We know that:
\[ x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = y^2 - 2 \]
Now, we can find \( x^4 + \frac{1}{x^4} \):
\[ x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = (y^2 - 2)^2 - 2 \]
Expanding this gives:
\[ (y^2 - 2)^2 - 2 = y^4 - 4y^2 + 4 - 2 = y^4 - 4y^2 + 2 \]

### Step 4: Set up the equation
Now we set this equal to 194:
\[ y^4 - 4y^2 + 2 = 194 \]
Subtracting 194 from both sides gives:
\[ y^4 - 4y^2 - 192 = 0 \]

### Step 5: Substitute \( z = y^2 \)
Let \( z = y^2 \). Then we have:
\[ z^2 - 4z - 192 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ z = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-192)}}{2 \cdot 1} \]
\[ z = \frac{4 \pm \sqrt{16 + 768}}{2} \]
\[ z = \frac{4 \pm \sqrt{784}}{2} \]
\[ z = \frac{4 \pm 28}{2} \]
Calculating the two possible values for \( z \):
1. \( z = \frac{32}{2} = 16 \)
2. \( z = \frac{-24}{2} = -12 \) (not valid since \( z \) must be non-negative)

Thus, \( z = 16 \).

### Step 7: Find \( y \)
Since \( z = y^2 \):
\[ y^2 = 16 \implies y = 4 \quad (\text{since } x > 0) \]

### Conclusion
Thus, the value of \( x + \frac{1}{x} \) is:
\[ \boxed{4} \]