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

(A) 1,2 (B) 1,0 (C) -1,2 (D) None of these

Given equation is $\begin{vmatrix}1 &4&20\\1&-2&5\\1&2x&5x^2\end{vmatrix}=0$
On expanding the determinant we get a quadratic equation in $x$.
It has two roots.
We observe that $R_3$ becomes identical to $R_1$ if $x=2$ thus at $x=2$
$\Rightarrow \Delta =0$
$\therefore x=2$ is a root of the equation .
Similarly $R_3$ becomes identical to $R_2$ if $x=-1$
At $x=-1$ is a root of the given equation .
Hence equation has roots as -1 and 2
Hence (c) is the correct answer.