Question

If $x=-9$ is a root of

$\left|\begin{array}{ccc}x& 3& 7\\ 2& x& 2\\ 7& 6& x\end{array}\right|=0$,
then other two roots are

Answer 1

Given: $\left|\begin{array}{ccc}x& 3& 7\\ 2& x& 2\\ 7& 6& x\end{array}\right|=0$,

$\Rightarrow \left|\begin{array}{ccc}x+9& x+9& x+9\\ 2& x& 2\\ 7& 6& x\end{array}\right|=0$

$[\text{Applying}{R}_1\to R_1+R_2+R_3]$

$\Rightarrow (x+9)\left|\begin{array}{ccc}1& 1& 1\\ 2& x& 2\\ 7& 6& x\end{array}\right|=0$,

$\left[\text{Taking}\right(x+9\left)\text{common from}{R}_1\right]$

$\Rightarrow (x+9)\left|\begin{array}{ccc}0& 0& 1\\ 2-x& x-2& 2\\ 1& 6-x& x\end{array}\right|=0$

$[\text{Applying}{C}_1\to C_1-C_2\text{and}{C}_2\to C_2-C_3]$

$\Rightarrow (x+9)\left[1\right((2-x)(6-x)-1(x-2)\left)\right]=0$

[Expanding along $R_1$]

$\Rightarrow (x+9)\left[\right(2-x\left)\right(6-x)+(2-x\left)\right]=0$

$\Rightarrow (x+9)\left[\right(2-x\left)\right(6-x+1\left)\right]=0$

$\Rightarrow (x+9)\left[\right(x-2\left)\right(7-x\left)\right]=0$

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

Given root is −9.

Hence, other two roots are 2 and 7.