If a,b are non zero real numbers and `alpha, beta` the roots of `x^(2)+ax+b=0`, then

Options

A. `alpha^(2),beta^(2)` are the roots of `x^(2)-(2b-a^(2))x+a^(2)=0`
B. `1/(alpha),1/(beta)` are the roots of `bx^(2)+ax+1=0`
C. `(alpha)/(beta),(beta)/(alpha)` are the roots of `bx^(2)+(2b-a^(2))x+b=0`
D. `(alpha-1),(beta-1)` are the roots of the equation `x^(2)+x(a+2)+1+a+b=0`

Answer

To solve the problem, we start with the given quadratic equation:

\[ x^2 + ax + b = 0 \]

### Step 1: Identify the roots
Let the roots of the equation be \( \alpha \) and \( \beta \). According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -a \)
- The product of the roots \( \alpha \beta = b \)

### Step 2: Analyze the options
We need to check the validity of the given options based on the relationships we derived from Vieta's formulas.

#### Option 1:
The product of the roots is \( \alpha^2 + \beta^2 \) and should equal \( a^2 \).
- We know \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \).
- Substituting the values, we get:
\[ \alpha^2 + \beta^2 = (-a)^2 - 2b = a^2 - 2b \]
- For this option to hold true, we would need \( a^2 - 2b = a^2 \), which implies \( b = 0 \). This contradicts the condition that \( b \) is non-zero.

#### Option 2:
The product of the roots is \( \frac{1}{b} \) and the sum is \( \frac{1}{\alpha} + \frac{1}{\beta} \).
- The product of the roots \( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{b} \) is correct.
- The sum of the roots \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-a}{b} \) is also correct.

#### Option 3:
The product of the roots is \( \frac{1}{b} \) and the sum is \( \frac{1}{\alpha + \beta} \).
- The product of the roots \( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{b} \) is correct.
- However, the sum \( \frac{1}{\alpha + \beta} = \frac{1}{-a} \) does not match \( \frac{-a}{b} \).

### Conclusion:
From the analysis, we find that:
- Option 1 is incorrect.
- Option 2 is correct.
- Option 3 is incorrect.

### Final Answer:
The correct option is **Option 2**.