The sum of two non - integral roots of
`Delta=|(x,3,4),(5,x,5),(4,2,x)|` = 0 is

Options

A. 4
B. `-4`
C. 16
D. none

Answer

To solve the given determinant problem, we need to find the sum of the two non-integral roots of the determinant \( \Delta = \begin{vmatrix} x & 3 & 4 \\ 5 & x & 5 \\ 4 & 2 & x \end{vmatrix} = 0 \).

### Step-by-Step Solution:

1. **Calculate the Determinant**:
We will first compute the determinant \( \Delta \):
\[ \Delta = \begin{vmatrix} x & 3 & 4 \\ 5 & x & 5 \\ 4 & 2 & x \end{vmatrix} \]

2. **Apply Row Operations**:
We can simplify the determinant by performing row operations. We will replace the first column \( C_1 \) with \( C_1 - C_3 \):
\[ C_1 \rightarrow C_1 - C_3 \]
This gives us:
\[ \Delta = \begin{vmatrix} x - 4 & 3 & 4 \\ 1 & x & 5 \\ 0 & 2 & x \end{vmatrix} \]

3. **Expand the Determinant**:
Now we can expand the determinant. We will use the first column:
\[ \Delta = (x - 4) \begin{vmatrix} x & 5 \\ 2 & x \end{vmatrix} - 3 \begin{vmatrix} 1 & 5 \\ 0 & x \end{vmatrix} + 4 \begin{vmatrix} 1 & x \\ 0 & 2 \end{vmatrix} \]

4. **Calculate the 2x2 Determinants**:
- For \( \begin{vmatrix} x & 5 \\ 2 & x \end{vmatrix} = x^2 - 10 \)
- For \( \begin{vmatrix} 1 & 5 \\ 0 & x \end{vmatrix} = x \)
- For \( \begin{vmatrix} 1 & x \\ 0 & 2 \end{vmatrix} = 2 \)

5. **Substitute Back**:
Substitute the values back into the determinant:
\[ \Delta = (x - 4)(x^2 - 10) - 3x + 8 \]

6. **Expand and Simplify**:
Expanding gives:
\[ \Delta = (x^3 - 10x + 4x^2 - 40) - 3x + 8 \]
Combine like terms:
\[ \Delta = x^3 + 4x^2 - 13x - 32 \]

7. **Set the Determinant to Zero**:
We need to find the roots of the equation:
\[ x^3 + 4x^2 - 13x - 32 = 0 \]

8. **Find the Roots**:
By using the Rational Root Theorem or synthetic division, we can find that \( x = 4 \) is a root. Thus, we can factor the polynomial:
\[ (x - 4)(x^2 + 8x + 8) = 0 \]

9. **Find the Non-Integral Roots**:
Now we need to find the roots of the quadratic \( x^2 + 8x + 8 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{64 - 32}}{2} = \frac{-8 \pm \sqrt{32}}{2} = \frac{-8 \pm 4\sqrt{2}}{2} = -4 \pm 2\sqrt{2} \]

10. **Sum of Non-Integral Roots**:
The two non-integral roots are \( -4 + 2\sqrt{2} \) and \( -4 - 2\sqrt{2} \). Their sum is:
\[ (-4 + 2\sqrt{2}) + (-4 - 2\sqrt{2}) = -8 \]

### Final Answer:
The sum of the two non-integral roots is \( -8 \).