If `x=-9` is a root of `|(x,3,7),(2,x,2),(7,6,x)|=0` then other two roots are…………………

Explanation

To solve the problem, we need to evaluate the determinant given by:

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

and find the other roots when \( x = -9 \) is one of the roots.

### Step 1: Calculate the Determinant

We will calculate the determinant using the first row:

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

### Step 2: Compute the 2x2 Determinants

1. For the first determinant:

\[ \begin{vmatrix} x & 2 \\ 6 & x \end{vmatrix} = x^2 - 12 \]

2. For the second determinant:

\[ \begin{vmatrix} 2 & 2 \\ 7 & x \end{vmatrix} = 2x - 14 \]

3. For the third determinant:

\[ \begin{vmatrix} 2 & x \\ 7 & 6 \end{vmatrix} = 12 - 7x \]

### Step 3: Substitute Back into the Determinant Expression

Now substituting these back into the determinant \( D \):

\[ D = x(x^2 - 12) - 3(2x - 14) + 7(12 - 7x) \]

### Step 4: Expand and Simplify

Expanding this expression:

\[ D = x^3 - 12x - 6x + 42 + 84 - 49x \]

Combining like terms:

\[ D = x^3 - 67x + 126 \]

### Step 5: Set the Determinant Equal to Zero

Since we know that \( x = -9 \) is a root, we can substitute \( x = -9 \) into the polynomial:

\[ (-9)^3 - 67(-9) + 126 = 0 \]

### Step 6: Factor the Polynomial

We can factor the polynomial \( x^3 - 67x + 126 \) knowing one root is \( x = -9 \). We can perform synthetic division or polynomial long division to find the other factors.

### Step 7: Find the Other Roots

Using synthetic division with \( x + 9 \):

1. Divide \( x^3 - 67x + 126 \) by \( x + 9 \).
2. This will yield a quadratic polynomial.

After performing the division, we find:

\[ x^3 - 67x + 126 = (x + 9)(x^2 - 9x + 14) \]

### Step 8: Solve the Quadratic Equation

Now we solve the quadratic equation \( x^2 - 9x + 14 = 0 \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 14}}{2 \cdot 1} \]

Calculating the discriminant:

\[ b^2 - 4ac = 81 - 56 = 25 \]

Thus,

\[ x = \frac{9 \pm 5}{2} \]

This gives us the roots:

\[ x = \frac{14}{2} = 7 \quad \text{and} \quad x = \frac{4}{2} = 2 \]

### Final Answer

The other two roots are \( x = 2 \) and \( x = 7 \).

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