The average of two numbers is 10 and the product of two numbers is 96. Find the two numbers.

Options

A. 3,32
B. 12,8
C. 1,96
D. 24,4

Answer

To solve the problem, we need to find two numbers based on the given average and product. Let's denote the two numbers as \( A \) and \( B \).

### Step 1: Set up the equations based on the average
The average of two numbers \( A \) and \( B \) is given as 10. The formula for the average is:

\[ \text{Average} = \frac{A + B}{2} \]

Setting this equal to 10, we have:

\[ \frac{A + B}{2} = 10 \]

Multiplying both sides by 2 to eliminate the fraction gives:

\[ A + B = 20 \quad \text{(Equation 1)} \]

### Step 2: Set up the equation based on the product
We are also given that the product of the two numbers is 96. This can be expressed as:

\[ A \times B = 96 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously
Now we have two equations:
1. \( A + B = 20 \)
2. \( A \times B = 96 \)

From Equation 1, we can express \( B \) in terms of \( A \):

\[ B = 20 - A \]

### Step 4: Substitute \( B \) in Equation 2
Now, substitute \( B \) in Equation 2:

\[ A \times (20 - A) = 96 \]

Expanding this gives:

\[ 20A - A^2 = 96 \]

Rearranging the equation leads to:

\[ A^2 - 20A + 96 = 0 \]

### Step 5: Solve the quadratic equation
Now, we can solve this quadratic equation using the quadratic formula, where \( a = 1, b = -20, c = 96 \):

\[ A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Calculating the discriminant:

\[ b^2 - 4ac = (-20)^2 - 4 \times 1 \times 96 = 400 - 384 = 16 \]

Now substituting back into the formula:

\[ A = \frac{20 \pm \sqrt{16}}{2} \]
\[ A = \frac{20 \pm 4}{2} \]

This gives us two possible values for \( A \):

1. \( A = \frac{24}{2} = 12 \)
2. \( A = \frac{16}{2} = 8 \)

### Step 6: Find corresponding values of \( B \)
Using \( A + B = 20 \):

1. If \( A = 12 \), then \( B = 20 - 12 = 8 \).
2. If \( A = 8 \), then \( B = 20 - 8 = 12 \).

Thus, the two numbers are \( 12 \) and \( 8 \).

### Final Answer:
The two numbers are \( 12 \) and \( 8 \).

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