find the value of x+y+z if `x^(2)+y^(2)+z^(2)=18` and `xy+yz+zx=9` .

Options

A. `9`
B. `3`
C. `6`
D. `8`

Answer

To find the value of \( x + y + z \) given the equations \( x^2 + y^2 + z^2 = 18 \) and \( xy + yz + zx = 9 \), we can use the identity that relates these quantities.

### Step-by-Step Solution:

1. **Recall the identity**: We use the identity:
\[ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \]

2. **Substitute the known values**:
- From the problem, we know:
- \( x^2 + y^2 + z^2 = 18 \)
- \( xy + yz + zx = 9 \)
- Substitute these values into the identity:
\[ (x + y + z)^2 = 18 + 2(9) \]

3. **Calculate \( 2(xy + yz + zx) \)**:
\[ 2(9) = 18 \]

4. **Combine the results**:
\[ (x + y + z)^2 = 18 + 18 = 36 \]

5. **Take the square root**:
\[ x + y + z = \sqrt{36} \]

6. **Find the value**:
\[ x + y + z = 6 \]

Thus, the value of \( x + y + z \) is \( 6 \).