`(x-(1)/(x))^(2)+8(x+(1)/(x))=29, x ne0`:

Options

A. `-5, 4`
B. `3, 2`
C. `1, 3`
D. `(3+sqrt5)/(2),(-11pm 3sqrt(13))/(2)`

Answer

To solve the equation \((x - \frac{1}{x})^2 + 8(x + \frac{1}{x}) = 29\), we will follow these steps:

### Step 1: Expand the equation
We start by expanding \((x - \frac{1}{x})^2\):
\[ (x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2} \]
Thus, the equation becomes:
\[ x^2 - 2 + \frac{1}{x^2} + 8(x + \frac{1}{x}) = 29 \]

### Step 2: Substitute \(y = x + \frac{1}{x}\)
Let \(y = x + \frac{1}{x}\). Then we can express \(\frac{1}{x^2}\) in terms of \(y\):
\[ \frac{1}{x^2} = y^2 - 2 \]
Now substituting \(y\) into the equation gives:
\[ (x^2 + \frac{1}{x^2}) = (y^2 - 2) \Rightarrow (y^2 - 2) - 2 + 8y = 29 \]
This simplifies to:
\[ y^2 + 8y - 33 = 0 \]

### Step 3: Solve the quadratic equation
Now we will solve the quadratic equation \(y^2 + 8y - 33 = 0\) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 8\), and \(c = -33\):
\[ y = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-33)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ y = \frac{-8 \pm \sqrt{64 + 132}}{2} = \frac{-8 \pm \sqrt{196}}{2} = \frac{-8 \pm 14}{2} \]

### Step 4: Find the values of \(y\)
Calculating the two possible values for \(y\):
1. \(y = \frac{6}{2} = 3\)
2. \(y = \frac{-22}{2} = -11\)

### Step 5: Solve for \(x\)
Now we will find \(x\) for both values of \(y\).

**For \(y = 3\)**:
\[ x + \frac{1}{x} = 3 \Rightarrow x^2 - 3x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{3 \pm \sqrt{5}}{2} \]

**For \(y = -11\)**:
\[ x + \frac{1}{x} = -11 \Rightarrow x^2 + 11x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-11 \pm \sqrt{117}}{2} = \frac{-11 \pm 3\sqrt{13}}{2} \]

### Final Results
Thus, the solutions for \(x\) are:
1. \(x = \frac{3 + \sqrt{5}}{2}\)
2. \(x = \frac{3 - \sqrt{5}}{2}\)
3. \(x = \frac{-11 + 3\sqrt{13}}{2}\)
4. \(x = \frac{-11 - 3\sqrt{13}}{2}\)