Which of the following are the two roots of the equation `(x^(2) + 2)^(2) + 8x^(2) = 6x (x^(2) + 2)`?

Options

A. `1 pm i`
B. `2 pm i`
C. `1 pm sqrt(2)`
D. `2 pm I sqrt(2)`

Answer

To solve the equation \((x^2 + 2)^2 + 8x^2 = 6x(x^2 + 2)\), we will follow these steps:

### Step 1: Rewrite the Equation
Start with the given equation:
\[ (x^2 + 2)^2 + 8x^2 = 6x(x^2 + 2) \]

### Step 2: Expand Both Sides
Expand the left side:
\[ (x^2 + 2)^2 = x^4 + 4x^2 + 4 \]
So, the left side becomes:
\[ x^4 + 4x^2 + 4 + 8x^2 = x^4 + 12x^2 + 4 \]

Now expand the right side:
\[ 6x(x^2 + 2) = 6x^3 + 12x \]

### Step 3: Set the Equation to Zero
Now, set the equation to zero:
\[ x^4 + 12x^2 + 4 - 6x^3 - 12x = 0 \]
Rearranging gives:
\[ x^4 - 6x^3 + 12x^2 - 12x + 4 = 0 \]

### Step 4: Substitute Variable
Let \(t = \frac{x^2 + 2}{x}\). Then, we can rewrite the equation in terms of \(t\):
\[ t^2 - 6t + 8 = 0 \]

### Step 5: Solve the Quadratic Equation
Now we will solve the quadratic equation \(t^2 - 6t + 8 = 0\) using the factorization method:
\[ (t - 4)(t - 2) = 0 \]
This gives us:
\[ t = 4 \quad \text{or} \quad t = 2 \]

### Step 6: Substitute Back for \(x\)
Now substitute back for \(t\):
1. For \(t = 4\):
\[ \frac{x^2 + 2}{x} = 4 \implies x^2 + 2 = 4x \implies x^2 - 4x + 2 = 0 \]
2. For \(t = 2\):
\[ \frac{x^2 + 2}{x} = 2 \implies x^2 + 2 = 2x \implies x^2 - 2x + 2 = 0 \]

### Step 7: Solve Each Quadratic Equation
1. For \(x^2 - 4x + 2 = 0\):
Using the quadratic formula:
\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = 2 \pm \sqrt{2} \]

2. For \(x^2 - 2x + 2 = 0\):
Using the quadratic formula:
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i \]

### Conclusion
The roots of the original equation are:
- From \(x^2 - 4x + 2 = 0\): \(2 \pm \sqrt{2}\)
- From \(x^2 - 2x + 2 = 0\): \(1 \pm i\)

Thus, the two roots of the equation are \(1 \pm i\).