For `alpha,beta in R`,suppose the system of linear equations
`x-y+z=5`
`2x+2y+alpha z=8`
`3x-y+4z=beta` has infinitely many solutions.Then `alpha` and `beta` are the roots of

Options

A. `x^(2)-18x+56=0`
B. `x^(2)+14x+24=0`
C. `x^(2)-10x+16=0`
D. `x^(2)+18x+56=0`

Answer

To solve the problem, we need to find the values of \(\alpha\) and \(\beta\) such that the system of equations has infinitely many solutions. This occurs when the determinant of the coefficients of the system is zero.

### Step 1: Write the system of equations in matrix form
The given equations are:
1. \(x - y + z = 5\)
2. \(2x + 2y + \alpha z = 8\)
3. \(3x - y + 4z = \beta\)

We can represent this system in matrix form as:
\[ \begin{bmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ \beta \end{bmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix
The determinant of the coefficient matrix must be zero for the system to have infinitely many solutions. The determinant \(\Delta\) can be calculated as follows:
\[ \Delta = \begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{vmatrix} \]

Calculating the determinant:
\[ \Delta = 1 \cdot \begin{vmatrix} 2 & \alpha \\ -1 & 4 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & \alpha \\ 3 & 4 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 2 \\ 3 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 2 & \alpha \\ -1 & 4 \end{vmatrix} = (2)(4) - (-1)(\alpha) = 8 + \alpha\)

2. \(\begin{vmatrix} 2 & \alpha \\ 3 & 4 \end{vmatrix} = (2)(4) - (3)(\alpha) = 8 - 3\alpha\)

3. \(\begin{vmatrix} 2 & 2 \\ 3 & -1 \end{vmatrix} = (2)(-1) - (3)(2) = -2 - 6 = -8\)

Putting it all together:
\[ \Delta = 1(8 + \alpha) + (8 - 3\alpha) - 8 \]
\[ \Delta = 8 + \alpha + 8 - 3\alpha - 8 = 8 - 2\alpha \]

### Step 3: Set the determinant to zero
For the system to have infinitely many solutions:
\[ 8 - 2\alpha = 0 \]
\[ 2\alpha = 8 \implies \alpha = 4 \]

### Step 4: Find \(\beta\) using the second determinant condition
Next, we need to find \(\beta\). We will use the determinant of the modified matrix where we replace the last column with the constants:
\[ \Delta_z = \begin{vmatrix} 1 & -1 & 5 \\ 2 & 2 & 8 \\ 3 & -1 & \beta \end{vmatrix} \]

Calculating this determinant:
\[ \Delta_z = 1 \cdot \begin{vmatrix} 2 & 8 \\ -1 & \beta \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & 8 \\ 3 & \beta \end{vmatrix} + 5 \cdot \begin{vmatrix} 2 & 2 \\ 3 & -1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 2 & 8 \\ -1 & \beta \end{vmatrix} = (2)(\beta) - (8)(-1) = 2\beta + 8\)

2. \(\begin{vmatrix} 2 & 8 \\ 3 & \beta \end{vmatrix} = (2)(\beta) - (8)(3) = 2\beta - 24\)

3. \(\begin{vmatrix} 2 & 2 \\ 3 & -1 \end{vmatrix} = -8\) (as calculated before)

Putting it all together:
\[ \Delta_z = 1(2\beta + 8) + (2\beta - 24) + 5(-8) \]
\[ \Delta_z = 2\beta + 8 + 2\beta - 24 - 40 \]
\[ \Delta_z = 4\beta - 56 \]

Setting this equal to zero:
\[ 4\beta - 56 = 0 \implies 4\beta = 56 \implies \beta = 14 \]

### Step 5: Form the quadratic equation
Now we have \(\alpha = 4\) and \(\beta = 14\). The roots of the quadratic equation can be formed as:
\[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \]
Substituting the values:
\[ x^2 - (4 + 14)x + (4 \cdot 14) = 0 \]
\[ x^2 - 18x + 56 = 0 \]

### Final Answer
Thus, the quadratic equation whose roots are \(\alpha\) and \(\beta\) is:
\[ \boxed{x^2 - 18x + 56 = 0} \]