If x = –9 is a root of

Solution:

Given: $x=-9$ is a root of $\left|\begin{array}{lll}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|=0$

Let $\Delta=\left|\begin{array}{lll}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|$

$\Delta=\left|\begin{array}{lll}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|$

Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$

$=\left|\begin{array}{ccc}x+2+7 & 3+x+6 & 7+2+x \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|$

$=\left|\begin{array}{ccc}x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|$

Taking out $(x+9)$ common from $R_{1}$

$=(x+9)\left|\begin{array}{lll}1 & 1 & 1 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|$

Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$

$=(x+9)\left|\begin{array}{lll}1 & 1-1 & 1-1 \\ 2 & x-2 & 2-2 \\ 7 & 6-7 & x-7\end{array}\right|$

$=(x+9)\left|\begin{array}{ccc}1 & 0 & 0 \\ 2 & x-2 & 0 \\ 7 & -1 & x-7\end{array}\right|$

Expanding through $R_{1}$

$=(x+9)(1(x-2)(x-7))$

$=(x+9)(x-2)(x-7)$

Since, $\Delta=0$

$\Rightarrow(x+9)(x-2)(x-7)=0$

$\Rightarrow x=-9,2,7$

Hence, other two roots are $\underline{2}$ and 7 .