The product of two numbers is 192 and the sum of these two numbers is 28. What is the smaller of these two numbers?

Options

A. 16
B. 14
C. 12
D. 18

Answer

To solve the problem, we need to find two numbers whose product is 192 and whose sum is 28. Let's denote these two numbers as \( n_1 \) and \( n_2 \).

### Step 1: Set up the equations
We have two equations based on the problem statement:
1. \( n_1 \times n_2 = 192 \) (Equation 1)
2. \( n_1 + n_2 = 28 \) (Equation 2)

### Step 2: Express one variable in terms of the other
From Equation 2, we can express \( n_2 \) in terms of \( n_1 \):
\[ n_2 = 28 - n_1 \]

### Step 3: Substitute into the first equation
Now, substitute \( n_2 \) into Equation 1:
\[ n_1 \times (28 - n_1) = 192 \]

### Step 4: Expand and rearrange the equation
Expanding the equation gives:
\[ 28n_1 - n_1^2 = 192 \]
Rearranging this into standard quadratic form:
\[ n_1^2 - 28n_1 + 192 = 0 \]

### Step 5: Solve the quadratic equation
To solve the quadratic equation \( n_1^2 - 28n_1 + 192 = 0 \), we can use the quadratic formula:
\[ n_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -28 \), and \( c = 192 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-28)^2 - 4 \times 1 \times 192 = 784 - 768 = 16 \]

Now applying the quadratic formula:
\[ n_1 = \frac{28 \pm \sqrt{16}}{2 \times 1} = \frac{28 \pm 4}{2} \]

Calculating the two possible values for \( n_1 \):
1. \( n_1 = \frac{28 + 4}{2} = \frac{32}{2} = 16 \)
2. \( n_1 = \frac{28 - 4}{2} = \frac{24}{2} = 12 \)

### Step 6: Identify the smaller number
The two numbers we found are \( n_1 = 16 \) and \( n_2 = 12 \). The smaller of these two numbers is:
\[ \text{Smaller number} = 12 \]

### Final Answer
The smaller of the two numbers is **12**.

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