The roots of the equations `|{:(1+x,3,5),(2,2+x,5),(2,3,x+4):}|` = 0 are

Options

A. 2,1,-9
B. 1,1,-9
C. `-1, 1,-9`
D. `-2,-1,-8`

Answer

To solve the determinant equation given by

\[ \begin{vmatrix} 1 + x & 3 & 5 \\ 2 & 2 + x & 5 \\ 2 & 3 & x + 4 \end{vmatrix} = 0, \]

we will follow these steps:

### Step 1: Write the determinant
We start with the determinant:

\[ D = \begin{vmatrix} 1 + x & 3 & 5 \\ 2 & 2 + x & 5 \\ 2 & 3 & x + 4 \end{vmatrix} \]

### Step 2: Apply row operations
We will simplify the determinant using row operations. Specifically, we can perform the following operations:

- \( R_2 \rightarrow R_2 - R_1 \)
- \( R_3 \rightarrow R_3 - R_1 \)

This gives us:

\[ D = \begin{vmatrix} 1 + x & 3 & 5 \\ 2 - (1 + x) & (2 + x) - 3 & 5 - 5 \\ 2 - (1 + x) & 3 - 3 & (x + 4) - 5 \end{vmatrix} \]

Calculating the new rows:

- For \( R_2 \):
- First element: \( 2 - (1 + x) = 1 - x \)
- Second element: \( (2 + x) - 3 = x - 1 \)
- Third element: \( 0 \)

- For \( R_3 \):
- First element: \( 2 - (1 + x) = 1 - x \)
- Second element: \( 0 \)
- Third element: \( (x + 4) - 5 = x - 1 \)

Thus, we have:

\[ D = \begin{vmatrix} 1 + x & 3 & 5 \\ 1 - x & x - 1 & 0 \\ 1 - x & 0 & x - 1 \end{vmatrix} \]

### Step 3: Factor out common terms
Notice that \( R_2 \) and \( R_3 \) both have a common factor of \( 1 - x \). We can factor this out:

\[ D = (1 - x) \begin{vmatrix} 1 + x & 3 & 5 \\ 1 & x - 1 & 0 \\ 1 & 0 & x - 1 \end{vmatrix} \]

### Step 4: Expand the determinant
Now we will expand the determinant:

\[ D = (1 - x) \begin{vmatrix} 1 + x & 3 & 5 \\ 1 & x - 1 & 0 \\ 1 & 0 & x - 1 \end{vmatrix} \]

Using the property of determinants, we can expand along the first column:

\[ D = (1 - x) \left( (1 + x) \begin{vmatrix} x - 1 & 0 \\ 0 & x - 1 \end{vmatrix} - 3 \begin{vmatrix} 1 & 0 \\ 1 & x - 1 \end{vmatrix} + 5 \begin{vmatrix} 1 & x - 1 \\ 1 & 0 \end{vmatrix} \right) \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} x - 1 & 0 \\ 0 & x - 1 \end{vmatrix} = (x - 1)(x - 1) = (x - 1)^2 \)

2. \( \begin{vmatrix} 1 & 0 \\ 1 & x - 1 \end{vmatrix} = 1(x - 1) - 0 = x - 1 \)

3. \( \begin{vmatrix} 1 & x - 1 \\ 1 & 0 \end{vmatrix} = 1(0) - 1(x - 1) = - (x - 1) \)

Substituting these back into the determinant:

\[ D = (1 - x) \left( (1 + x)(x - 1)^2 - 3(x - 1) - 5(x - 1) \right) \]

### Step 5: Simplify the expression
Combining the terms:

\[ D = (1 - x) \left( (1 + x)(x - 1)^2 - 8(x - 1) \right) \]

### Step 6: Set the determinant to zero
Setting \( D = 0 \):

\[ (1 - x) \left( (1 + x)(x - 1)^2 - 8(x - 1) \right) = 0 \]

This gives us two cases:

1. \( 1 - x = 0 \) → \( x = 1 \)
2. \( (1 + x)(x - 1)^2 - 8(x - 1) = 0 \)

### Step 7: Solve the second equation
Factoring out \( (x - 1) \):

\[ (x - 1) \left( (1 + x)(x - 1) - 8 \right) = 0 \]

This leads to:

\[ (1 + x)(x - 1) - 8 = 0 \]

Expanding gives:

\[ x^2 - 1 + x - 8 = 0 \implies x^2 + x - 9 = 0 \]

### Step 8: Use the quadratic formula
Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 36}}{2} = \frac{-1 \pm 7}{2} \]

This gives us:

\[ x = 3 \quad \text{and} \quad x = -4 \]

### Final Roots
Thus, the roots of the equation are:

\[ x = 1, \quad x = 3, \quad x = -4 \]