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 given problem step-by-step, we will follow these steps:

### Step 1: Identify the roots of the quadratic equation
The given quadratic equation is:
\[ x^2 - 2x + 3 = 0 \]

Using the quadratic formula, the roots \( \alpha \) and \( \beta \) can be calculated as:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -2, c = 3 \).

### Step 2: Calculate the sum and product of the roots
From Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = 2 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = 3 \)

### Step 3: Calculate the new expressions for the roots
We need to find the values of:
\[ \alpha^3 - 3\alpha^2 + 5\alpha - 2 \]
and
\[ \beta^3 - \beta^2 + \beta + 5 \]

### Step 4: Simplify the expressions
1. **For \( \alpha \)**:
\[ \alpha^3 - 3\alpha^2 + 5\alpha - 2 \]
We can express \( \alpha^3 \) in terms of \( \alpha \) and \( \beta \):
Using the original quadratic equation, we have:
\[ \alpha^2 = 2\alpha - 3 \implies \alpha^3 = \alpha \cdot \alpha^2 = \alpha(2\alpha - 3) = 2\alpha^2 - 3\alpha \]
Substituting \( \alpha^2 \):
\[ = 2(2\alpha - 3) - 3\alpha = 4\alpha - 6 - 3\alpha = \alpha - 6 \]
Therefore:
\[ \alpha^3 - 3\alpha^2 + 5\alpha - 2 = (\alpha - 6) - 3(2\alpha - 3) + 5\alpha - 2 \]
Simplifying:
\[ = \alpha - 6 - 6\alpha + 9 + 5\alpha - 2 = 0 \]

2. **For \( \beta \)**:
\[ \beta^3 - \beta^2 + \beta + 5 \]
Similarly, we can express \( \beta^3 \):
\[ \beta^3 = \beta(2\beta - 3) = 2\beta^2 - 3\beta \]
Substituting \( \beta^2 \):
\[ = 2(2\beta - 3) - 3\beta = 4\beta - 6 - 3\beta = \beta - 6 \]
Therefore:
\[ \beta^3 - \beta^2 + \beta + 5 = (\beta - 6) - (2\beta - 3) + \beta + 5 \]
Simplifying:
\[ = \beta - 6 - 2\beta + 3 + \beta + 5 = 2 \]

### Step 5: Calculate the sum of the new roots
Now we have:
- The first root \( \alpha^3 - 3\alpha^2 + 5\alpha - 2 = 0 \)
- The second root \( \beta^3 - \beta^2 + \beta + 5 = 2 \)

Thus, the sum of the new roots is:
\[ 0 + 2 = 2 \]

### Final Answer
The sum of the roots of the new equation is:
\[ \boxed{2} \]