Solve the equation `x^3 -6x^2 +7x +2=0` one root being ` 2 + sqrt(5)`

Answer

To solve the equation \( x^3 - 6x^2 + 7x + 2 = 0 \) given that one root is \( 2 + \sqrt{5} \), we can follow these steps:

### Step 1: Identify the given root
We know that one root of the equation is \( r_1 = 2 + \sqrt{5} \).

### Step 2: Find the conjugate root
Since the coefficients of the polynomial are real numbers, the conjugate of the given root must also be a root. Thus, the second root is:
\[ r_2 = 2 - \sqrt{5} \]

### Step 3: Let the third root be \( r_3 \)
Let the third root be \( r_3 = \alpha \).

### Step 4: Use Vieta's formulas
According to Vieta's formulas, the sum of the roots of the polynomial \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ r_1 + r_2 + r_3 = -\frac{b}{a} \]
For our equation, \( a = 1 \) and \( b = -6 \). Therefore:
\[ r_1 + r_2 + r_3 = -\frac{-6}{1} = 6 \]

### Step 5: Substitute the known roots
Substituting the known roots into the equation:
\[ (2 + \sqrt{5}) + (2 - \sqrt{5}) + \alpha = 6 \]

### Step 6: Simplify the equation
The terms \( \sqrt{5} \) and \( -\sqrt{5} \) cancel out:
\[ 2 + 2 + \alpha = 6 \]
This simplifies to:
\[ 4 + \alpha = 6 \]

### Step 7: Solve for \( \alpha \)
Now, solving for \( \alpha \):
\[ \alpha = 6 - 4 = 2 \]

### Step 8: List all roots
Thus, the roots of the equation are:
1. \( r_1 = 2 + \sqrt{5} \)
2. \( r_2 = 2 - \sqrt{5} \)
3. \( r_3 = 2 \)

### Final Answer
The roots of the equation \( x^3 - 6x^2 + 7x + 2 = 0 \) are:
\[ 2 + \sqrt{5}, \quad 2 - \sqrt{5}, \quad 2 \]

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