If `alpha+beta=3` and `a^(3)+beta^(3) = 7`, then `alpha` and `beta` are the roots of the equation

Options

A. `9x^(2) + 27 x + 20 = 0`
B. `9x^(2) - 27 x + 20 = 0`
C. `9x^(2) + 27 x - 20 = 0`
D. `9x^(2) - 27 x - 20 = 0`

Answer

To solve the problem step by step, we will use the given information about the roots \( \alpha \) and \( \beta \).

### Step 1: Use the given information
We know:
- \( \alpha + \beta = 3 \)
- \( \alpha^3 + \beta^3 = 7 \)

### Step 2: Apply the identity for the sum of cubes
We can use the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We can also express \( \alpha^2 + \beta^2 \) in terms of \( \alpha + \beta \) and \( \alpha\beta \):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Thus, we can rewrite the sum of cubes as:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ \alpha^3 + \beta^3 = 3 \left(3^2 - 3\alpha\beta\right) = 3(9 - 3\alpha\beta) \]
Setting this equal to 7:
\[ 3(9 - 3\alpha\beta) = 7 \]

### Step 4: Solve for \( \alpha\beta \)
Now, we simplify the equation:
\[ 27 - 9\alpha\beta = 7 \]
\[ 9\alpha\beta = 27 - 7 \]
\[ 9\alpha\beta = 20 \]
\[ \alpha\beta = \frac{20}{9} \]

### Step 5: Form the quadratic equation
The roots \( \alpha \) and \( \beta \) can be used to form the quadratic equation:
\[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \]
Substituting the values we found:
\[ x^2 - 3x + \frac{20}{9} = 0 \]

### Step 6: Clear the fraction
To eliminate the fraction, multiply the entire equation by 9:
\[ 9x^2 - 27x + 20 = 0 \]

### Step 7: Identify the correct option
The quadratic equation is:
\[ 9x^2 - 27x + 20 = 0 \]
Thus, the roots \( \alpha \) and \( \beta \) satisfy this equation.

### Final Answer
The roots \( \alpha \) and \( \beta \) are the roots of the equation:
\[ 9x^2 - 27x + 20 = 0 \]

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