Two numbers whose sum is 12 and the absolute value of whose difference is 4 are the roots of the equation____

Options

A. `x^(2)- 12x + 30 = 0`
B. `x^(2)- 12x+ 32= 0`
C. `2x^(2)- 6x+ 7= 0`
D. `2x^(2) - 24x+ 43= 0`

Answer

To find the quadratic equation whose roots are two numbers whose sum is 12 and the absolute value of whose difference is 4, we can follow these steps:

### Step 1: Define the Variables
Let the two numbers be \( \alpha \) and \( \beta \).

### Step 2: Set Up the Equations
From the problem, we know:
1. The sum of the numbers:
\[ \alpha + \beta = 12 \]
2. The absolute value of the difference of the numbers:
\[ |\alpha - \beta| = 4 \]
This gives us two cases:
- Case 1: \( \alpha - \beta = 4 \)
- Case 2: \( \beta - \alpha = 4 \)

### Step 3: Solve the Equations
#### Case 1: \( \alpha - \beta = 4 \)
We can add the two equations:
\[ \alpha + \beta = 12 \quad (1) \]
\[ \alpha - \beta = 4 \quad (2) \]

Adding (1) and (2):
\[ (\alpha + \beta) + (\alpha - \beta) = 12 + 4 \]
\[ 2\alpha = 16 \]
\[ \alpha = 8 \]

Now substitute \( \alpha = 8 \) back into equation (1):
\[ 8 + \beta = 12 \]
\[ \beta = 12 - 8 = 4 \]

So, in this case, the two numbers are \( \alpha = 8 \) and \( \beta = 4 \).

#### Case 2: \( \beta - \alpha = 4 \)
We can also solve this case similarly:
\[ \alpha + \beta = 12 \quad (1) \]
\[ \beta - \alpha = 4 \quad (3) \]

Adding (1) and (3):
\[ (\alpha + \beta) + (\beta - \alpha) = 12 + 4 \]
\[ 2\beta = 16 \]
\[ \beta = 8 \]

Now substitute \( \beta = 8 \) back into equation (1):
\[ \alpha + 8 = 12 \]
\[ \alpha = 12 - 8 = 4 \]

So, in this case, the two numbers are \( \alpha = 4 \) and \( \beta = 8 \).

### Step 4: Form the Quadratic Equation
The roots of the quadratic equation can be expressed using the formula:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]

Substituting \( \alpha + \beta = 12 \) and \( \alpha \beta = 8 \times 4 = 32 \):
\[ x^2 - 12x + 32 = 0 \]

### Final Answer
The quadratic equation whose roots are the two numbers is:
\[ x^2 - 12x + 32 = 0 \]