Square of difference between two numbers is 9 while the sum of squares of those two numbers is 225. What is their product?

Options

A. a. 108
B. b. 125
C. c. 169
D. d. 96

Answer

To solve the problem, we need to find the product of two numbers \( x \) and \( y \) given the following conditions:

1. The square of the difference between the two numbers is 9:
\[ (x - y)^2 = 9 \]

2. The sum of the squares of the two numbers is 225:
\[ x^2 + y^2 = 225 \]

### Step 1: Solve for \( x - y \)

From the first condition, we can take the square root of both sides:
\[ x - y = 3 \quad \text{or} \quad x - y = -3 \]

### Step 2: Express \( x \) in terms of \( y \)

Let's consider the case \( x - y = 3 \):
\[ x = y + 3 \]

### Step 3: Substitute into the second condition

Now, substitute \( x \) into the second condition:
\[ x^2 + y^2 = 225 \]
Substituting \( x = y + 3 \):
\[ (y + 3)^2 + y^2 = 225 \]

### Step 4: Expand and simplify

Expanding the equation:
\[ (y^2 + 6y + 9) + y^2 = 225 \]
Combine like terms:
\[ 2y^2 + 6y + 9 = 225 \]

### Step 5: Rearrange the equation

Rearranging gives:
\[ 2y^2 + 6y + 9 - 225 = 0 \]
\[ 2y^2 + 6y - 216 = 0 \]

### Step 6: Simplify the quadratic equation

Divide the entire equation by 2:
\[ y^2 + 3y - 108 = 0 \]

### Step 7: Factor the quadratic equation

Now we need to factor the quadratic:
\[ (y + 12)(y - 9) = 0 \]

### Step 8: Solve for \( y \)

Setting each factor to zero gives:
\[ y + 12 = 0 \quad \Rightarrow \quad y = -12 \]
\[ y - 9 = 0 \quad \Rightarrow \quad y = 9 \]

### Step 9: Find corresponding \( x \) values

For \( y = 9 \):
\[ x = y + 3 = 9 + 3 = 12 \]

For \( y = -12 \):
\[ x = y + 3 = -12 + 3 = -9 \]

### Step 10: Calculate the product \( xy \)

Now, we can find the product \( xy \):
For \( x = 12 \) and \( y = 9 \):
\[ xy = 12 \times 9 = 108 \]

For \( x = -9 \) and \( y = -12 \):
\[ xy = -9 \times -12 = 108 \]

Thus, in both cases, the product of the two numbers is:
\[ \boxed{108} \]