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 determine the nature of its roots, we will follow these steps:

### Step 1: Substitute \( t = x^2 \)
We start by substituting \( t = x^2 \). This transforms the original equation into a quadratic form:
\[ t^2 - 2t + 4 = 0 \]

### Step 2: Use the quadratic formula
Next, we apply the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots of the quadratic equation. Here, \( a = 1 \), \( b = -2 \), and \( c = 4 \):
\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
\[ t = \frac{2 \pm \sqrt{4 - 16}}{2} \]
\[ t = \frac{2 \pm \sqrt{-12}}{2} \]

### Step 3: Simplify the square root
We simplify the square root of the negative number:
\[ \sqrt{-12} = \sqrt{12} \cdot i = 2\sqrt{3}i \]
Thus, we can rewrite \( t \):
\[ t = \frac{2 \pm 2\sqrt{3}i}{2} = 1 \pm \sqrt{3}i \]

### Step 4: Find \( x \) from \( t \)
Since \( t = x^2 \), we have:
\[ x^2 = 1 + \sqrt{3}i \quad \text{and} \quad x^2 = 1 - \sqrt{3}i \]

### Step 5: Find the square roots
To find \( x \), we need to take the square root of both expressions. We can express these in polar form. First, we find the modulus and argument for \( 1 + \sqrt{3}i \):
- Modulus:
\[ |1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \]
- Argument:
\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]
Thus, we can write:
\[ 1 + \sqrt{3}i = 2 \text{cis} \frac{\pi}{3} \]
where \( \text{cis} \theta = \cos \theta + i \sin \theta \).

Taking the square root:
\[ x = \sqrt{2} \text{cis} \frac{\pi}{6} \quad \text{or} \quad x = \sqrt{2} \text{cis} \left(\frac{\pi}{6} + \pi\right) \]
This gives us:
\[ x = \sqrt{2} \text{cis} \frac{\pi}{6} \quad \text{and} \quad x = \sqrt{2} \text{cis} \frac{7\pi}{6} \]

For \( 1 - \sqrt{3}i \):
- Modulus:
\[ |1 - \sqrt{3}i| = 2 \]
- Argument:
\[ \theta = \tan^{-1}\left(-\sqrt{3}\right) = -\frac{\pi}{3} \]
Thus:
\[ 1 - \sqrt{3}i = 2 \text{cis} \left(-\frac{\pi}{3}\right) \]
Taking the square root:
\[ x = \sqrt{2} \text{cis} \left(-\frac{\pi}{6}\right) \quad \text{or} \quad x = \sqrt{2} \text{cis} \left(-\frac{\pi}{6} + \pi\right) \]
This gives us:
\[ x = \sqrt{2} \text{cis} \left(-\frac{\pi}{6}\right) \quad \text{and} \quad x = \sqrt{2} \text{cis} \left(\frac{5\pi}{6}\right) \]

### Step 6: Summary of the roots
Thus, the roots of the original equation are:
1. \( \sqrt{2} \text{cis} \frac{\pi}{6} \)
2. \( \sqrt{2} \text{cis} \frac{7\pi}{6} \)
3. \( \sqrt{2} \text{cis} \left(-\frac{\pi}{6}\right) \)
4. \( \sqrt{2} \text{cis} \frac{5\pi}{6} \)

### Conclusion
The roots represent the vertices of a rectangle inscribed in a circle of radius \( \sqrt{2} \).

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