If one root of the equation `|{:(7,6,x^(2)-13),(2,x^(2)-13,2),(x^(2)-13,3,7):}|` =0 is x=2 the sum of all other five roots is

Options

A. `2sqrt(15)`
B. -2
C. `sqrt(20)+sqrt(15)-2`
D. None of these

Answer

To solve the given problem, we need to evaluate the determinant and find the roots of the resulting polynomial equation. Let's break this down step by step.

### Step 1: Set up the determinant
We are given the determinant:

\[ D = \begin{vmatrix} 7 & 6 & x^2 - 13 \\ 2 & x^2 - 13 & 2 \\ x^2 - 13 & 3 & 7 \end{vmatrix} \]

### Step 2: Calculate the determinant
To calculate the determinant, we can use the formula for a 3x3 determinant:

\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where:
- \( a = 7, b = 6, c = x^2 - 13 \)
- \( d = 2, e = x^2 - 13, f = 2 \)
- \( g = x^2 - 13, h = 3, i = 7 \)

Calculating each part:

1. \( ei - fh = (x^2 - 13) \cdot 7 - 2 \cdot 3 = 7x^2 - 91 - 6 = 7x^2 - 97 \)

2. \( di - fg = 2 \cdot 7 - 2(x^2 - 13) = 14 - 2x^2 + 26 = 40 - 2x^2 \)

3. \( dh - eg = 2 \cdot 3 - (x^2 - 13)(x^2 - 13) = 6 - (x^2 - 13)^2 \)

Now substituting back into the determinant formula:

\[ D = 7(7x^2 - 97) - 6(40 - 2x^2) + (x^2 - 13)(6 - (x^2 - 13)^2) \]

### Step 3: Simplify the determinant
Now we simplify \( D \):

\[ D = 49x^2 - 679 - 240 + 12x^2 + (x^2 - 13)(6 - (x^4 - 26x^2 + 169)) \]

We need to expand and combine like terms to obtain a polynomial in \( x \).

### Step 4: Set the determinant equal to zero
Since we know that one root is \( x = 2 \), we can substitute \( x = 2 \) into the polynomial to verify it is a root and then find the sum of the other roots.

### Step 5: Use Vieta's formulas
The polynomial will be of degree 6, and by Vieta's formulas, the sum of the roots of the polynomial \( ax^6 + bx^5 + ... = 0 \) is given by \( -\frac{b}{a} \).

### Step 6: Determine the sum of the other roots
Since \( x = 2 \) is one of the roots, we can denote the sum of the other five roots as \( S \). Thus, we have:

\[ 2 + S = -\frac{b}{a} \]

From the previous calculations, if \( b = 0 \) (since there is no \( x^5 \) term), then:

\[ 2 + S = 0 \implies S = -2 \]

### Final Answer
The sum of all other five roots is:

\[ \boxed{-2} \]