If one of the roots of the equation `|(7,6,x^(2)-25),(2,x^(2)-25,2),(x^(2)-25,3,7)|=0` is `x=3`, then the sum of all other five roots is

Options

A. 0
B. `-3`
C. `-6`
D. `-8`

Answer

To solve the given problem, we need to evaluate the determinant and find the sum of the roots based on the information provided. Let's break it down step by step.

### Step 1: Set Up the Determinant
We are given the determinant:
\[ D = \begin{vmatrix} 7 & 6 & x^2 - 25 \\ 2 & x^2 - 25 & 2 \\ x^2 - 25 & 3 & 7 \end{vmatrix} \]
and it is stated that one of the roots is \(x = 3\).

### Step 2: Substitute \(x = 3\) into the Determinant
First, we substitute \(x = 3\) into the determinant:
\[ D = \begin{vmatrix} 7 & 6 & 3^2 - 25 \\ 2 & 3^2 - 25 & 2 \\ 3^2 - 25 & 3 & 7 \end{vmatrix} \]
Calculating \(3^2 - 25\):
\[ 3^2 - 25 = 9 - 25 = -16 \]
Thus, the determinant becomes:
\[ D = \begin{vmatrix} 7 & 6 & -16 \\ 2 & -16 & 2 \\ -16 & 3 & 7 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Now we will calculate the determinant using cofactor expansion along the first row:
\[ D = 7 \begin{vmatrix} -16 & 2 \\ 3 & 7 \end{vmatrix} - 6 \begin{vmatrix} 2 & 2 \\ -16 & 7 \end{vmatrix} + (-16) \begin{vmatrix} 2 & -16 \\ -16 & 3 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -16 & 2 \\ 3 & 7 \end{vmatrix} = (-16)(7) - (2)(3) = -112 - 6 = -118\)

2. \(\begin{vmatrix} 2 & 2 \\ -16 & 7 \end{vmatrix} = (2)(7) - (2)(-16) = 14 + 32 = 46\)

3. \(\begin{vmatrix} 2 & -16 \\ -16 & 3 \end{vmatrix} = (2)(3) - (-16)(-16) = 6 - 256 = -250\)

Now substituting back into the determinant:
\[ D = 7(-118) - 6(46) - 16(-250) \]
Calculating each term:
- \(7(-118) = -826\)
- \(-6(46) = -276\)
- \(-16(-250) = 4000\)

Now combine these:
\[ D = -826 - 276 + 4000 = 4000 - 1102 = 2898 \]

### Step 4: Set the Determinant Equal to Zero
Since we know that the determinant equals zero for the roots:
\[ D = 0 \]
This means we need to find the polynomial formed by the determinant and its roots.

### Step 5: Find the Sum of the Roots
The polynomial formed by the determinant is of degree 6. The sum of the roots of a polynomial can be found using Vieta's formulas. The sum of the roots is given by:
\[ -\frac{b}{a} \]
where \(b\) is the coefficient of \(x^{n-1}\) and \(a\) is the leading coefficient.

In our case, since the \(x^5\) term is missing, we have:
\[ b = 0 \quad \text{and} \quad a = 1 \quad \Rightarrow \quad \text{Sum of all roots} = -\frac{0}{1} = 0 \]

### Step 6: Calculate the Sum of the Other Roots
We know one root is \(x = 3\). Therefore, the sum of the other 5 roots is:
\[ \text{Sum of other roots} = 0 - 3 = -3 \]

### Final Answer
The sum of all other five roots is \(-3\).