If `x^(2)+xy+xz=18,y^(2)+yz+yx+12=0` and `z^(2)+zx+zy=30`, then `(x+y+z)=`

Options

A. 6
B. 3
C. `-2`
D. `-3`

Answer

To solve the problem, we have the following equations:

1. \( x^2 + xy + xz = 18 \) (Equation 1)
2. \( y^2 + yz + yx + 12 = 0 \) (Equation 2)
3. \( z^2 + zx + zy = 30 \) (Equation 3)

We need to find the value of \( x + y + z \).

### Step 1: Rearranging Equation 2
From Equation 2, we can rearrange it to isolate the quadratic terms:
\[ y^2 + yz + yx = -12 \]
This means:
\[ y^2 + y(z + x) = -12 \]

### Step 2: Adding the Equations
Now, we will add all three equations together:
\[ (x^2 + xy + xz) + (y^2 + yz + yx) + (z^2 + zx + zy) = 18 + (-12) + 30 \]
This simplifies to:
\[ x^2 + y^2 + z^2 + xy + xz + yz + yx + zx + zy = 36 \]

### Step 3: Simplifying the Left Side
Notice that \( xy + yx = 2xy \), \( xz + zx = 2zx \), and \( yz + zy = 2yz \). Therefore, we can rewrite the left side as:
\[ x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 36 \]
This can be recognized as:
\[ (x + y + z)^2 = 36 \]

### Step 4: Taking the Square Root
Now, we take the square root of both sides:
\[ x + y + z = \pm 6 \]

### Step 5: Final Answer
Since we are looking for the value of \( x + y + z \), we can conclude:
\[ x + y + z = 6 \quad \text{(considering the positive root)} \]

Thus, the final answer is:
\[ \boxed{6} \]