Question

The roots of the equation $\left| \begin{matrix} 1 & 4 & 20 \\ 1 & - 2 & 5 \\ 1 & 2x & 5x^{2} \end{matrix} \right| = 0$are.

• A. $- 1, - 2$
• B. $- 1,2$
• C. $1, - 2$
• D. $1,2$

Answer 1

Answer: (B)

$- 1,2$

Answer 2

$\left| \begin{matrix} 1 & 4 & 20 \\ 1 & - 2 & 5 \\ 1 & 2x & 5x^{2} \end{matrix} \right| = 0$

$\Rightarrow$ $\left| \begin{matrix} 0 & 6 & 15 \\ 0 & - 2 - 2x & 5(1 - x^{2}) \\ 1 & 2x & 5x^{2} \end{matrix} \right| = 0$ $\left( \begin{aligned} & R_{1} \rightarrow R_{1} - R_{2} \\ & R_{2} \rightarrow R_{2} - R_{3} \end{aligned} \right)$

0 & 1 & 1 \\ 0 & - (1 + x) & 1 - x^{2} \\ 1 & x & x^{2} \end{matrix} \right| = 0$$ $\Rightarrow$ $(1 + x)\left| \begin{matrix} 0 & 1 & 1 \\ 0 & - 1 & 1 - x \\ 1 & x & x^{2} \end{matrix} \right| = 0$ $\Rightarrow$ $x + 1 = 0$ or $x - 2 = 0$ $\Rightarrow$ $x = - 1,2$. **Trick:** Obviously by inspection, $x = - 1,2$ satisfy the equation. At $x = - 1$ $\left| \begin{matrix} 1 & 4 & 20 \\ 1 & - 2 & 5 \\ 1 & - 2 & 5 \end{matrix} \right| = 0$ as $R_{2} \equiv R_{3}$ At $x = 2$, $\left| \begin{matrix} 1 & 4 & 20 \\ 1 & - 2 & 5 \\ 1 & 4 & 20 \end{matrix} \right| = 0$ as $R_{1} \equiv R_{3}$.