The operation & is defined as `r & s =(s^(2)-r^(2))/(r+s)` where r and s are real number and `r ne -s`. What is the value of 3 & 4?

Answer

To find the value of \( 3 \& 4 \) using the defined operation \( r \& s = \frac{s^2 - r^2}{r + s} \), we will follow these steps:

### Step 1: Identify the values of \( r \) and \( s \)
Here, we have:
- \( r = 3 \)
- \( s = 4 \)

### Step 2: Substitute \( r \) and \( s \) into the operation
We substitute the values into the operation:
\[ 3 \& 4 = \frac{4^2 - 3^2}{3 + 4} \]

### Step 3: Calculate \( s^2 \) and \( r^2 \)
Calculate \( 4^2 \) and \( 3^2 \):
\[ 4^2 = 16 \quad \text{and} \quad 3^2 = 9 \]

### Step 4: Substitute the squares back into the equation
Now substitute these values back into the equation:
\[ 3 \& 4 = \frac{16 - 9}{3 + 4} \]

### Step 5: Simplify the numerator and denominator
Calculate the numerator and denominator:
\[ 16 - 9 = 7 \quad \text{and} \quad 3 + 4 = 7 \]
So we have:
\[ 3 \& 4 = \frac{7}{7} \]

### Step 6: Simplify the fraction
Now simplify the fraction:
\[ \frac{7}{7} = 1 \]

### Final Answer
Thus, the value of \( 3 \& 4 \) is:
\[ \boxed{1} \]