Determine real values of x and y for which each statement is true
`(x^(4) + 2x i)- (3x^(2) + yi)= (3-5i) + (1 + 2yi)`

Answer

To solve the equation \((x^{4} + 2xi) - (3x^{2} + yi) = (3 - 5i) + (1 + 2yi)\), we will separate the real and imaginary parts and solve for \(x\) and \(y\).

### Step 1: Rewrite the equation
We start with the equation:
\[ (x^{4} + 2xi) - (3x^{2} + yi) = (3 - 5i) + (1 + 2yi) \]
This simplifies to:
\[ x^{4} + 2xi - 3x^{2} - yi = 4 - 5i + 2yi \]

### Step 2: Separate real and imaginary parts
Now, we can separate the real and imaginary parts:
- Real part: \(x^{4} - 3x^{2} = 4\)
- Imaginary part: \(2x - y = -5 + 2y\)

### Step 3: Set up equations
From the real part, we have:
\[ x^{4} - 3x^{2} - 4 = 0 \tag{1} \]
From the imaginary part, we rearrange to get:
\[ 2x - y = -5 + 2y \implies 2x + 5 = 3y \implies y = \frac{2x + 5}{3} \tag{2} \]

### Step 4: Solve the polynomial equation (1)
Let \(t = x^{2}\). Then, we rewrite equation (1):
\[ t^{2} - 3t - 4 = 0 \]
We can factor this as:
\[ (t - 4)(t + 1) = 0 \]
Thus, we have:
\[ t - 4 = 0 \quad \text{or} \quad t + 1 = 0 \]
This gives us:
\[ t = 4 \quad \text{or} \quad t = -1 \]
Since \(t = x^{2}\) and must be non-negative, we discard \(t = -1\) and take:
\[ x^{2} = 4 \implies x = 2 \quad \text{or} \quad x = -2 \]

### Step 5: Find corresponding \(y\) values using (2)
1. For \(x = 2\):
\[ y = \frac{2(2) + 5}{3} = \frac{4 + 5}{3} = \frac{9}{3} = 3 \]

2. For \(x = -2\):
\[ y = \frac{2(-2) + 5}{3} = \frac{-4 + 5}{3} = \frac{1}{3} \]

### Final Answer
Thus, the real values of \(x\) and \(y\) are:
- When \(x = 2\), \(y = 3\)
- When \(x = -2\), \(y = \frac{1}{3}\)