If `alpha and beta` are roots of the quadratic equation `x ^(2) + 4x +3=0,` then the equation whose roots are `2 alpha + beta and alpha + 2 beta` is :

Options

A. `x ^(2) -12x +35=0`
B. `x ^(2)+ 12x-33 =0`
C. `x ^(2) -12x-33=0`
D. `x ^(2) +12x+35=0`

Answer

To solve the problem, we need to find the new quadratic equation whose roots are \(2\alpha + \beta\) and \(\alpha + 2\beta\), given that \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2 + 4x + 3 = 0\).

### Step-by-step Solution:

1. **Identify the roots of the original equation**:
The given quadratic equation is:
\[ x^2 + 4x + 3 = 0 \]
To find the roots, we can factor the equation:
\[ (x + 1)(x + 3) = 0 \]
Thus, the roots are:
\[ \alpha = -1 \quad \text{and} \quad \beta = -3 \]

2. **Calculate the new roots**:
We need to find the new roots \(2\alpha + \beta\) and \(\alpha + 2\beta\):
\[ 2\alpha + \beta = 2(-1) + (-3) = -2 - 3 = -5 \]
\[ \alpha + 2\beta = -1 + 2(-3) = -1 - 6 = -7 \]

3. **Sum of the new roots**:
The sum of the new roots \(\alpha_1\) and \(\beta_1\) is:
\[ \alpha_1 + \beta_1 = (-5) + (-7) = -12 \]

4. **Product of the new roots**:
The product of the new roots is:
\[ \alpha_1 \beta_1 = (-5)(-7) = 35 \]

5. **Form the new quadratic equation**:
The standard form of a quadratic equation with roots \(r_1\) and \(r_2\) is:
\[ x^2 - (r_1 + r_2)x + r_1 r_2 = 0 \]
Substituting the values we found:
\[ x^2 - (-12)x + 35 = 0 \]
This simplifies to:
\[ x^2 + 12x + 35 = 0 \]

### Final Answer:
The new quadratic equation is:
\[ \boxed{x^2 + 12x + 35 = 0} \]