If `alpha` and `beta` the roots of the equation `x^(2)-2x+3=0`, then the sum of roots of the equation having roots as `alpha^(3)-3alpha^(2)+5alpha-2` and `beta^(3)-beta^(2)+beta+5` is

Options

A. 1
B. 3
C. 5
D. 7

Answer

To solve the problem step by step, we need to find the sum of the roots of the new equations formed by the expressions given for α and β, which are the roots of the equation \(x^2 - 2x + 3 = 0\).

### Step 1: Identify the roots α and β
The roots of the equation \(x^2 - 2x + 3 = 0\) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -2\), and \(c = 3\).

Calculating the discriminant:
\[ b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8 \]
Since the discriminant is negative, the roots are complex:
\[ x = \frac{2 \pm \sqrt{-8}}{2} = 1 \pm i\sqrt{2} \]
Thus, we have:
\[ \alpha = 1 + i\sqrt{2}, \quad \beta = 1 - i\sqrt{2} \]

### Step 2: Calculate the expressions for the roots
We need to evaluate:
1. \( \alpha^3 - 3\alpha^2 + 5\alpha - 2 \)
2. \( \beta^3 - \beta^2 + \beta + 5 \)

#### For \( \alpha \):
Using the equation \( \alpha^2 - 2\alpha + 3 = 0 \), we can express \( \alpha^2 \) as:
\[ \alpha^2 = 2\alpha - 3 \]
Now, to find \( \alpha^3 \):
\[ \alpha^3 = \alpha \cdot \alpha^2 = \alpha(2\alpha - 3) = 2\alpha^2 - 3\alpha \]
Substituting \( \alpha^2 \):
\[ \alpha^3 = 2(2\alpha - 3) - 3\alpha = 4\alpha - 6 - 3\alpha = \alpha - 6 \]
Now substituting back into the expression:
\[ \alpha^3 - 3\alpha^2 + 5\alpha - 2 = (\alpha - 6) - 3(2\alpha - 3) + 5\alpha - 2 \]
Expanding this:
\[ = \alpha - 6 - 6\alpha + 9 + 5\alpha - 2 = 0 \]

#### For \( \beta \):
Similarly, we can find \( \beta^3 \):
\[ \beta^3 = \beta(2\beta - 3) = 2\beta^2 - 3\beta \]
Substituting \( \beta^2 \):
\[ \beta^3 = 2(2\beta - 3) - 3\beta = 4\beta - 6 - 3\beta = \beta - 6 \]
Now substituting back into the expression:
\[ \beta^3 - \beta^2 + \beta + 5 = (\beta - 6) - (2\beta - 3) + \beta + 5 \]
Expanding this:
\[ = \beta - 6 - 2\beta + 3 + \beta + 5 = 0 \]

### Step 3: Find the sum of the roots of the new equation
Since both expressions evaluate to 0, the new roots are 0 and 0. Therefore, the sum of the roots is:
\[ 0 + 0 = 0 \]

### Final Answer
The sum of the roots of the equation having roots as \( \alpha^3 - 3\alpha^2 + 5\alpha - 2 \) and \( \beta^3 - \beta^2 + \beta + 5 \) is \( \boxed{0} \).