The roots of the equation `(3 - x)^(4) + (2 - x)^(4) = (5 - 2x)^(4)` are

Options

A. all real
B. all imaginary
C. two real and two imaginary
D. none of these

Answer

To solve the equation \((3 - x)^{4} + (2 - x)^{4} = (5 - 2x)^{4}\), we will follow these steps:

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

### Step 2: Rewrite the equation
Substituting \(x\) into the original equation gives:
\[ (3 - \frac{5 - t}{2})^{4} + (2 - \frac{5 - t}{2})^{4} = t^{4} \]
Simplifying the terms:
\[ (3 - \frac{5}{2} + \frac{t}{2})^{4} + (2 - \frac{5}{2} + \frac{t}{2})^{4} = t^{4} \]
This simplifies to:
\[ (\frac{1}{2} + \frac{t}{2})^{4} + (-\frac{1}{2} + \frac{t}{2})^{4} = t^{4} \]

### Step 3: Simplify further
Now we can rewrite the equation:
\[ \left(\frac{t + 1}{2}\right)^{4} + \left(\frac{t - 1}{2}\right)^{4} = t^{4} \]
Expanding both terms:
\[ \frac{(t + 1)^{4}}{16} + \frac{(t - 1)^{4}}{16} = t^{4} \]
Multiplying through by 16 to eliminate the fraction:
\[ (t + 1)^{4} + (t - 1)^{4} = 16t^{4} \]

### Step 4: Expand both sides
Expanding \((t + 1)^{4}\) and \((t - 1)^{4}\):
\[ (t + 1)^{4} = t^{4} + 4t^{3} + 6t^{2} + 4t + 1 \]
\[ (t - 1)^{4} = t^{4} - 4t^{3} + 6t^{2} - 4t + 1 \]
Adding these:
\[ (t + 1)^{4} + (t - 1)^{4} = 2t^{4} + 12t^{2} + 2 \]
Setting this equal to \(16t^{4}\):
\[ 2t^{4} + 12t^{2} + 2 = 16t^{4} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 14t^{4} - 12t^{2} + 2 = 0 \]
Dividing through by 2:
\[ 7t^{4} - 6t^{2} + 1 = 0 \]

### Step 6: Let \(u = t^{2}\)
Let \(u = t^{2}\), then we have:
\[ 7u^{2} - 6u + 1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\):
\[ u = \frac{6 \pm \sqrt{(-6)^{2} - 4 \cdot 7 \cdot 1}}{2 \cdot 7} \]
Calculating the discriminant:
\[ u = \frac{6 \pm \sqrt{36 - 28}}{14} = \frac{6 \pm \sqrt{8}}{14} = \frac{6 \pm 2\sqrt{2}}{14} = \frac{3 \pm \sqrt{2}}{7} \]

### Step 8: Finding \(t\)
Since \(u = t^{2}\), we have:
\[ t^{2} = \frac{3 + \sqrt{2}}{7} \quad \text{or} \quad t^{2} = \frac{3 - \sqrt{2}}{7} \]
Taking square roots:
\[ t = \pm \sqrt{\frac{3 + \sqrt{2}}{7}} \quad \text{and} \quad t = \pm \sqrt{\frac{3 - \sqrt{2}}{7}} \]

### Step 9: Finding \(x\)
Recall \(t = 5 - 2x\):
\[ 5 - 2x = \sqrt{\frac{3 + \sqrt{2}}{7}} \quad \text{or} \quad 5 - 2x = -\sqrt{\frac{3 + \sqrt{2}}{7}} \]
Solving for \(x\):
\[ x = \frac{5 - \sqrt{\frac{3 + \sqrt{2}}{7}}}{2} \quad \text{and} \quad x = \frac{5 + \sqrt{\frac{3 + \sqrt{2}}{7}}}{2} \]
And similarly for the other case.

### Final Roots
Thus, we find two real roots and two imaginary roots from the calculations.