The roots of the equation `x^(4)-2x^(2)+4=0` are the vertices of `a` :

Options

A. square inscribed in a circle of radius `2`
B. rectangle inscribed in a circle of radius `2`
C. square inscribed in a circle of radius `sqrt(2)`
D. rectangle inscribed in a circle of radius `sqrt(2)`

Explanation

To solve the equation \( x^4 - 2x^2 + 4 = 0 \) and find the roots, we can follow these steps:

### Step 1: Substitute \( x^2 \) with \( t \)
Let \( t = x^2 \). Then the equation becomes:
\[ t^2 - 2t + 4 = 0 \]

### Step 2: Identify coefficients
In the quadratic equation \( at^2 + bt + c = 0 \), we have:
- \( a = 1 \)
- \( b = -2 \)
- \( c = 4 \)

### Step 3: Use the quadratic formula
The quadratic formula is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
This simplifies to:
\[ t = \frac{2 \pm \sqrt{4 - 16}}{2} \]
\[ t = \frac{2 \pm \sqrt{-12}}{2} \]

### Step 4: Simplify the square root
Since \( \sqrt{-12} = \sqrt{12}i = 2\sqrt{3}i \), we can write:
\[ t = \frac{2 \pm 2\sqrt{3}i}{2} \]
This simplifies to:
\[ t = 1 \pm \sqrt{3}i \]

### Step 5: Solve for \( x \)
Recall that \( t = x^2 \). Therefore, we have:
\[ x^2 = 1 + \sqrt{3}i \quad \text{and} \quad x^2 = 1 - \sqrt{3}i \]

### Step 6: Find the square roots
To find \( x \), we need to compute the square roots of the complex numbers.

1. For \( x^2 = 1 + \sqrt{3}i \):
- Convert to polar form:
- Magnitude: \( r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2 \)
- Argument: \( \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \)
- Thus, \( x^2 = 2 \left( \cos\frac{\pi}{3} + i \sin\frac{\pi}{3} \right) \)
- Taking the square root:
\[ x = \sqrt{2} \left( \cos\frac{\pi/3}{2} + i \sin\frac{\pi/3}{2} \right) = \sqrt{2} \left( \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} \right) \]
- This gives:
\[ x = \sqrt{2} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \quad \text{and} \quad x = -\sqrt{2} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \]

2. For \( x^2 = 1 - \sqrt{3}i \):
- Similarly, convert to polar form:
- Magnitude: \( r = 2 \)
- Argument: \( \theta = \tan^{-1}\left(-\frac{\sqrt{3}}{1}\right) = -\frac{\pi}{3} \)
- Thus, \( x^2 = 2 \left( \cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right) \right) \)
- Taking the square root:
\[ x = \sqrt{2} \left( \cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right) \right) \]
- This gives:
\[ x = \sqrt{2} \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) \quad \text{and} \quad x = -\sqrt{2} \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) \]

### Step 7: List the roots
The four roots are:
1. \( x_1 = \sqrt{2} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \)
2. \( x_2 = -\sqrt{2} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \)
3. \( x_3 = \sqrt{2} \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) \)
4. \( x_4 = -\sqrt{2} \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) \)

### Conclusion
The roots \( x_1, x_2, x_3, x_4 \) represent the vertices of a rectangle in the complex plane.

---