Find the quadratic equation whose roots are `3+sqrt2, 3- sqrt2`.

Answer

To find the quadratic equation whose roots are \(3 + \sqrt{2}\) and \(3 - \sqrt{2}\), we can follow these steps:

### Step 1: Identify the roots
The roots given are:
- \(r_1 = 3 + \sqrt{2}\)
- \(r_2 = 3 - \sqrt{2}\)

### Step 2: Calculate the sum of the roots
The sum of the roots \(S\) is given by:
\[ S = r_1 + r_2 = (3 + \sqrt{2}) + (3 - \sqrt{2}) \]
Simplifying this, we have:
\[ S = 3 + \sqrt{2} + 3 - \sqrt{2} = 6 \]

### Step 3: Calculate the product of the roots
The product of the roots \(P\) is given by:
\[ P = r_1 \cdot r_2 = (3 + \sqrt{2})(3 - \sqrt{2}) \]
Using the difference of squares formula \(a^2 - b^2\), we can simplify:
\[ P = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7 \]

### Step 4: Form the quadratic equation
The standard form of a quadratic equation with roots \(r_1\) and \(r_2\) is:
\[ x^2 - Sx + P = 0 \]
Substituting the values of \(S\) and \(P\):
\[ x^2 - 6x + 7 = 0 \]

### Final Answer
Thus, the quadratic equation whose roots are \(3 + \sqrt{2}\) and \(3 - \sqrt{2}\) is:
\[ \boxed{x^2 - 6x + 7 = 0} \]