If `alpha, beta` are the roots of `x^(2)+3x+1=0`, then the equation whose roots `2-alpha, 2-beta` is

Options

A. `x^(2) + 7x +10 = 0`
B. `x^(2) - 7x +11 = 0`
C. `x^(2) + 4x +11 = 0`
D. `x^(2) + 4x +10 = 0`

Answer

To find the equation whose roots are \(2 - \alpha\) and \(2 - \beta\) given that \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 + 3x + 1 = 0\), we can follow these steps:

### Step 1: Identify the coefficients of the original quadratic equation
The given equation is:
\[ x^2 + 3x + 1 = 0 \]
Here, we can identify:
- \(a = 1\)
- \(b = 3\)
- \(c = 1\)

### Step 2: Use Vieta's formulas to find the sum and product of the roots
According to Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{3}{1} = -3\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{1}{1} = 1\)

### Step 3: Find the sum of the new roots \(2 - \alpha\) and \(2 - \beta\)
The sum of the new roots can be calculated as follows:
\[ (2 - \alpha) + (2 - \beta) = 4 - (\alpha + \beta) = 4 - (-3) = 4 + 3 = 7 \]

### Step 4: Find the product of the new roots \(2 - \alpha\) and \(2 - \beta\)
The product of the new roots can be calculated as follows:
\[ (2 - \alpha)(2 - \beta) = 4 - 2(\alpha + \beta) + \alpha \beta \]
Substituting the values we found:
\[ = 4 - 2(-3) + 1 = 4 + 6 + 1 = 11 \]

### Step 5: Form the new quadratic equation
Using the sum and product of the new roots, we can form the new quadratic equation:
\[ x^2 - (\text{sum of the roots})x + (\text{product of the roots}) = 0 \]
Substituting the values:
\[ x^2 - 7x + 11 = 0 \]

### Final Answer
Thus, the equation whose roots are \(2 - \alpha\) and \(2 - \beta\) is:
\[ \boxed{x^2 - 7x + 11 = 0} \]