Explanation

(b)
Applying $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$ and taking common $(2x + 10)$ from $R_{1},$we get
$(2x + 10)\left| \begin{matrix}
1 & 1 & 1 \\
2 & 2x & 2 \\
7 & 6 & 2x
\end{matrix} \right| = 0$
$\Rightarrow (2x + 10)\left| \begin{matrix}
1 & 0 & 0 \\
2 & 2x - 2 & 0 \\
7 & - 1 & 2x - 7
\end{matrix} \right| = 0$
$\lbrack C_{3} \rightarrow C_{3} - C_{1}$and$\ C_{2} \rightarrow C_{2} - C_{1}\rbrack$
$\Rightarrow \ \ (2x + 10)(2x - 2)(2x - 7) = 0$
$\Rightarrow x = - 5,\ 1,\ \frac{7}{2}$
Hence , other roots are 1 and$\ \frac{7}{2}$ or 1 and 3.5