If `x` is root of `(1+x^(4))=7(1+x)^(4)` then `|x+1/x|=` ?

Explanation

To solve the equation \(1 + x^4 = 7(1 + x)^4\) and find \(|x + \frac{1}{x}|\), we will follow these steps:

### Step 1: Simplify the equation
Start with the given equation:
\[ 1 + x^4 = 7(1 + x)^4 \]

### Step 2: Expand the right-hand side
Expand \((1 + x)^4\) using the binomial theorem:
\[ (1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4 \]
Thus, we have:
\[ 1 + x^4 = 7(1 + 4x + 6x^2 + 4x^3 + x^4) \]

### Step 3: Distribute the 7
Distributing the 7 gives:
\[ 1 + x^4 = 7 + 28x + 42x^2 + 28x^3 + 7x^4 \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:
\[ x^4 - 7x^4 - 42x^2 - 28x^3 - 28x + 1 - 7 = 0 \]
This simplifies to:
\[ -6x^4 - 28x^3 - 42x^2 - 28x - 6 = 0 \]
Dividing through by -6 gives:
\[ x^4 + \frac{14}{3}x^3 + 7x^2 + \frac{14}{3}x + 1 = 0 \]

### Step 5: Substitute \(t = x + \frac{1}{x}\)
Let \(t = x + \frac{1}{x}\). Then, we can express \(x^2 + \frac{1}{x^2}\) in terms of \(t\):
\[ x^2 + \frac{1}{x^2} = t^2 - 2 \]

### Step 6: Substitute back into the equation
Substituting \(x + \frac{1}{x} = t\) into the rearranged equation:
\[ t^2 - 2 = 7(t + 2) \]
Expanding gives:
\[ t^2 - 2 = 7t + 14 \]
Rearranging leads to:
\[ t^2 - 7t - 16 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 49 + 64 = 113 \]
Thus:
\[ t = \frac{7 \pm \sqrt{113}}{2} \]

### Step 8: Find \(|x + \frac{1}{x}|\)
Now, we need to find \(|t|\):
The possible values for \(t\) are:
\[ t_1 = \frac{7 + \sqrt{113}}{2}, \quad t_2 = \frac{7 - \sqrt{113}}{2} \]
Since \(\sqrt{113} \approx 10.63\), we can evaluate:
\[ t_1 \approx \frac{7 + 10.63}{2} \approx 8.815, \quad t_2 \approx \frac{7 - 10.63}{2} \approx -1.815 \]
Thus, \(|t|\) can be:
\[ |t_1| \approx 8.815 \quad \text{and} \quad |t_2| \approx 1.815 \]

### Conclusion
The possible values for \(|x + \frac{1}{x}|\) are:
\[ |x + \frac{1}{x}| = \frac{7 + \sqrt{113}}{2} \quad \text{or} \quad |x + \frac{1}{x}| = \frac{7 - \sqrt{113}}{2} \]