If a, b, c are non - zero real numbers, the system of equations
`y+z=a+2x, x+z=b+2y, x+y=c+2z` is consistent and `b=4a+(c )/(4)`, then the sum of the roots of the equation `at^(2)+bt+c=0` is

Options

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

Answer

To solve the problem step by step, we will analyze the given system of equations and the condition provided.

### Step 1: Write the system of equations
The given system of equations is:
1. \( y + z = a + 2x \)
2. \( x + z = b + 2y \)
3. \( x + y = c + 2z \)

### Step 2: Rearranging the equations
We can rearrange these equations to express them in standard form:
1. \( 2x - y - z = -a \) (Equation 1)
2. \( x - 2y + z = b \) (Equation 2)
3. \( x + y - 2z = c \) (Equation 3)

### Step 3: Formulate the coefficient matrix and determinant
The coefficient matrix for this system is:
\[ \begin{bmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{bmatrix} \]
Let \( D \) be the determinant of this matrix. For the system to be consistent, we need \( D \neq 0 \) or \( D = 0 \) with the corresponding determinants of the modified matrices also being zero.

### Step 4: Calculate the determinant \( D \)
Calculating the determinant \( D \):
\[ D = \begin{vmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{vmatrix} \]
Using the determinant formula:
\[ D = 2 \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} - (-1) \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} = (-2)(-2) - (1)(1) = 4 - 1 = 3 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} = (1)(-2) - (1)(1) = -2 - 1 = -3 \)
3. \( \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-2)(1) = 1 + 2 = 3 \)

Putting it all together:
\[ D = 2(3) + 3 + 3 = 6 + 3 + 3 = 12 \]

### Step 5: Check for consistency
Since \( D \neq 0 \), the system is consistent.

### Step 6: Use the given condition \( b = 4a + \frac{c}{4} \)
We are given that \( b = 4a + \frac{c}{4} \). We can express this in terms of \( a, b, c \):
\[ 4b = 16a + c \]
This gives us the equation:
\[ 16a - 4b + c = 0 \quad (1) \]

### Step 7: Find the relationship among \( a, b, c \)
From the determinants calculated, we know:
\[ a + b + c = 0 \quad (2) \]

### Step 8: Solve the equations (1) and (2)
We now have two equations:
1. \( 16a - 4b + c = 0 \)
2. \( a + b + c = 0 \)

Substituting \( c = -a - b \) from equation (2) into equation (1):
\[ 16a - 4b - a - b = 0 \]
This simplifies to:
\[ 15a - 5b = 0 \implies 3a = b \implies b = 3a \]

### Step 9: Substitute back to find \( c \)
Using \( b = 3a \) in equation (2):
\[ a + 3a + c = 0 \implies 4a + c = 0 \implies c = -4a \]

### Step 10: Find the sum of the roots of the quadratic equation
The quadratic equation is given by:
\[ at^2 + bt + c = 0 \]
Substituting \( b = 3a \) and \( c = -4a \):
\[ at^2 + 3at - 4a = 0 \]
Dividing through by \( a \) (since \( a \neq 0 \)):
\[ t^2 + 3t - 4 = 0 \]

### Step 11: Calculate the sum of the roots
The sum of the roots of the quadratic equation \( at^2 + bt + c = 0 \) is given by:
\[ -\frac{b}{a} = -\frac{3a}{a} = -3 \]

### Conclusion
Thus, the sum of the roots of the equation \( at^2 + bt + c = 0 \) is:
\[ \boxed{-3} \]