Question

The solution set of the equation
$\begin{vmatrix}1&4&20\\ 1&-2&5\\ 1&2x&5x^{2}\end{vmatrix} = 0$ is

• A. {0, 1}
• B. {1, 2}
• C. {1, 5}
• D. {2, - 1}

Answer 1

Answer: (D)

Given $\begin{vmatrix}1&4&20\\ 1&-2&5\\ 1&2x&5x^{2}\end{vmatrix} = 0$
Operate, $ R_{2} \to R_{2} - R_{1} R_{3} \to R_{3} - R_{1} $
$\begin{vmatrix}1&4&20\\ 0&-6&-15\\ 0&2x-4&5x^{2}-20\end{vmatrix}=0$
$\Rightarrow 1\left[-30x^{2} + 120 + 30x -60\right] =0 $
$\Rightarrow 30x^{2} -30x -60 =0$
$ \Rightarrow x^{2} -x -2 =0$
$ \Rightarrow \left(x -2\right)\left(x+1\right)=0$
$ \Rightarrow x=-1,2 $
Thus, solution set is {2, - 1}.