Explanation

(b)
We have, $\left| \begin{matrix}
1 & 4 & 20 \\
1 & - 2 & 5 \\
1 & 2x & 5x^{2}
\end{matrix} \right| = 0$
$\Rightarrow \ \left| \begin{matrix}
0 & 6 & 15 \\
0 & - 2 - 2x & 5(1 - x^{2}) \\
1 & 2x & 5x^{2}
\end{matrix} \right| = 0\ \ \binom{R_{1} \rightarrow R_{1} - R_{2}}{and\ R_{2} \rightarrow R_{2} - R_{3}}$
$\Rightarrow 3 \bullet 2 \bullet 5\left| \begin{matrix}
0 & 1 & 1 \\
0 & - (1 + x) & 1 - x^{2} \\
1 & x & x^{2}
\end{matrix} \right| = 0$
(Taking common, 3 from $R_{1},$ 2 from $C_{2}$ , 5 from $C_{3}$)
$\Rightarrow (1 + x)\left| \begin{matrix}
0 & 1 & 1 \\
0 & - 1 & 1 - x \\
1 & x & x^{2}
\end{matrix} \right| = 0$
$\Rightarrow (1 + x)(2 - x) = 0$
$\Rightarrow x + 1 = 0\ or\ \ x - 2 = 0\ \Rightarrow x = - 1,\ 2$