If ` a+ b = 7 and ab= 6, ` find ` a^(2) - b^(2)`

Answer

To solve the problem where \( a + b = 7 \) and \( ab = 6 \), and we need to find \( a^2 - b^2 \), we can follow these steps:

### Step 1: Use the identity for \( a^2 - b^2 \)
We know that:
\[ a^2 - b^2 = (a + b)(a - b) \]
From the problem, we already have \( a + b = 7 \).

### Step 2: Find \( a - b \)
To find \( a - b \), we can use the equations we have. We know:
\[ ab = 6 \]
We can express \( b \) in terms of \( a \):
\[ b = 7 - a \]
Substituting this into the equation \( ab = 6 \):
\[ a(7 - a) = 6 \]
This simplifies to:
\[ 7a - a^2 = 6 \]
Rearranging gives us:
\[ a^2 - 7a + 6 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:
\[ a^2 - 7a + 6 = (a - 1)(a - 6) = 0 \]
Setting each factor to zero gives us:
\[ a - 1 = 0 \quad \text{or} \quad a - 6 = 0 \]
Thus, we find:
\[ a = 1 \quad \text{or} \quad a = 6 \]

### Step 4: Find corresponding values of \( b \)
Using \( b = 7 - a \):
- If \( a = 1 \), then \( b = 7 - 1 = 6 \).
- If \( a = 6 \), then \( b = 7 - 6 = 1 \).

### Step 5: Calculate \( a^2 - b^2 \)
Now we can calculate \( a^2 - b^2 \):
Using \( a = 1 \) and \( b = 6 \):
\[ a^2 - b^2 = 1^2 - 6^2 = 1 - 36 = -35 \]

Using \( a = 6 \) and \( b = 1 \):
\[ a^2 - b^2 = 6^2 - 1^2 = 36 - 1 = 35 \]

### Conclusion
Thus, the value of \( a^2 - b^2 \) is \( 35 \).