Q. If one root of determinant$\left| \begin{array}{r} x\ \ 3\ \ 7 \\ 2\ \ x\ \ 2 \\ 7\ \ 6\ \ x \end{array} \right| = 0,\$is$- 9,\$then the other two roots are
A) 2,7
B) 2,$- 7$
C) $- 2,7$
D) $- 2, - 7$
Explanation
(a)
Given, $\left| \begin{array}{r}
x\ \ 3\ \ 7 \\
2\ \ x\ \ 2 \\
7\ \ 6\ \ x
\end{array} \right|$=0
$\Rightarrow x\left( x^{2} - 12 \right) - 3(2x - 14) + 7(12 - 7x) = 0$
$\Rightarrow x^{3} - 67x + 126 = 0$
$\Rightarrow (x + 9)\left( x^{2} - 9x + 14 \right) = 0$
$\Rightarrow (x + 9)(x - 2)(x - 7) = 0$
$\Rightarrow x = - 9,\ 2,\ 7$
Hence, the other two roots are 2, 7
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.