Question

If x = -9 is a root of A = $\begin{vmatrix} x & 3 & 7 \\\ 2 & x & 2 \\\ 7 & 6 & x \\\ \end{vmatrix}$ = 0, then other two root are

• A. 3,7
• B. 2,7
• C. 3,6
• D. 2,6

Answer 1

Answer: (B)

2,7

Answer 2

GivenA = $\begin{vmatrix} x & 3 & 7 \\\ 2 & x & 2 \\\ 7 & 6 & x \\\ \end{vmatrix}$ = 0
$\Rightarrow$ x[x - 12] - 3[2x- 14] + 7[12 - 7x] = 0
$\Rightarrow \ x^3$ - 67x + 126 = 0
But given (x = 9) is a root of given determinant
$\therefore \ \ (x+9)$is a factor
$\Rightarrow \ \ x^3+9x^2-9x^2 -$81x + 14x + 126 = 0
$\Rightarrow \ \ \ x^2$(x + 9) - 9x(x + 9) + 14(x + 9) = 0
$\Rightarrow \ \ (x \ + \ 9) (x^2$-9x +14)=0
$\Rightarrow \ \ (x \ + \ 9) (x^2$-7x -2x+14)=0
$\Rightarrow$ (x + 9) (x - 7) (x - 2) = 0
$\Rightarrow \ \ x \ = \ -9,7,2$