Solve the equation
` x^4 -6x^3 +18x^2 - 30 x+ 25 =0` given that 2+i is a root

Answer

To solve the equation \( x^4 - 6x^3 + 18x^2 - 30x + 25 = 0 \) given that \( 2+i \) is a root, we can follow these steps:

### Step 1: Identify the roots
Since \( 2+i \) is a root and the coefficients of the polynomial are real, the complex conjugate \( 2-i \) must also be a root. Therefore, we have two roots:
- \( x_1 = 2 + i \)
- \( x_2 = 2 - i \)

### Step 2: Use Vieta's formulas
The polynomial is a quartic equation, meaning it has four roots. Let the other two roots be \( x_3 \) and \( x_4 \). According to Vieta's formulas, the sum of the roots \( x_1 + x_2 + x_3 + x_4 \) is equal to \( -\frac{b}{a} \), where \( b \) is the coefficient of \( x^3 \) and \( a \) is the coefficient of \( x^4 \).

For our polynomial:
\[ x_1 + x_2 + x_3 + x_4 = -\frac{-6}{1} = 6 \]
Substituting \( x_1 \) and \( x_2 \):
\[ (2+i) + (2-i) + x_3 + x_4 = 6 \]
This simplifies to:
\[ 4 + x_3 + x_4 = 6 \]
Thus, we find:
\[ x_3 + x_4 = 6 - 4 = 2 \]

### Step 3: Calculate the product of the roots
The product of the roots \( x_1 \cdot x_2 \cdot x_3 \cdot x_4 \) is equal to \( \frac{d}{a} \), where \( d \) is the constant term.
\[ x_1 \cdot x_2 \cdot x_3 \cdot x_4 = \frac{25}{1} = 25 \]
Calculating \( x_1 \cdot x_2 \):
\[ (2+i)(2-i) = 4 - i^2 = 4 + 1 = 5 \]
Thus:
\[ 5 \cdot x_3 \cdot x_4 = 25 \]
This gives:
\[ x_3 \cdot x_4 = \frac{25}{5} = 5 \]

### Step 4: Set up a system of equations
Now we have a system of equations:
1. \( x_3 + x_4 = 2 \) (Equation 1)
2. \( x_3 \cdot x_4 = 5 \) (Equation 2)

### Step 5: Solve for \( x_3 \) and \( x_4 \)
From Equation 1, we can express \( x_4 \) in terms of \( x_3 \):
\[ x_4 = 2 - x_3 \]
Substituting this into Equation 2:
\[ x_3(2 - x_3) = 5 \]
Expanding this gives:
\[ 2x_3 - x_3^2 = 5 \]
Rearranging leads to:
\[ x^2 - 2x + 5 = 0 \]

### Step 6: Use the quadratic formula
Now we can solve for \( x_3 \):
\[ x_3 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i \]
Thus, we have:
\[ x_3 = 1 + 2i \quad \text{and} \quad x_4 = 1 - 2i \]

### Final Roots
The roots of the equation are:
- \( 2 + i \)
- \( 2 - i \)
- \( 1 + 2i \)
- \( 1 - 2i \)