If 5 and 7 are the roots of the equation `|(x,4,5),(7,x,7),(5,8,x)| = 0`, then what is the third root ?

Options

A. -12
B. 9
C. 13
D. 14

Answer

To find the third root of the equation given by the determinant \( |(x,4,5),(7,x,7),(5,8,x)| = 0 \), where 5 and 7 are already known roots, we can follow these steps:

### Step 1: Calculate the Determinant
We start by calculating the determinant:

\[ D = \begin{vmatrix} x & 4 & 5 \\ 7 & x & 7 \\ 5 & 8 & x \end{vmatrix} \]

Using the formula for a 3x3 determinant, we expand along the first row:

\[ D = x \begin{vmatrix} x & 7 \\ 8 & x \end{vmatrix} - 4 \begin{vmatrix} 7 & 7 \\ 5 & x \end{vmatrix} + 5 \begin{vmatrix} 7 & x \\ 5 & 8 \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:

1. \( \begin{vmatrix} x & 7 \\ 8 & x \end{vmatrix} = x^2 - 56 \)

2. \( \begin{vmatrix} 7 & 7 \\ 5 & x \end{vmatrix} = 7x - 35 \)

3. \( \begin{vmatrix} 7 & x \\ 5 & 8 \end{vmatrix} = 56 - 5x \)

### Step 3: Substitute Back into the Determinant
Now substituting these back into our expression for \( D \):

\[ D = x(x^2 - 56) - 4(7x - 35) + 5(56 - 5x) \]

Expanding this gives:

\[ D = x^3 - 56x - 28x + 140 + 280 - 25x \]

### Step 4: Combine Like Terms
Combining like terms, we have:

\[ D = x^3 - 109x + 420 \]

### Step 5: Factor the Polynomial
We know that 5 and 7 are roots, so we can express \( D \) as:

\[ D = (x - 5)(x - 7)(x - r) \]

where \( r \) is the third root we need to find. The polynomial can also be expressed as:

\[ D = x^3 - (5 + 7 + r)x^2 + (5 \cdot 7 + 5r + 7r)x - 5 \cdot 7 \cdot r \]

### Step 6: Set Up the Equations
From the coefficients, we can set up the following equations:

1. \( 5 + 7 + r = 12 + r \) (coefficient of \( x^2 \))
2. \( 5 \cdot 7 + 5r + 7r = 35 + 12r \) (coefficient of \( x \))
3. \( 5 \cdot 7 \cdot r = 35r \) (constant term)

### Step 7: Solve for the Third Root
From the first equation, we can express \( r \):

\[ r = -12 \]

### Conclusion
Thus, the third root of the equation is:

\[ \boxed{-12} \]