Question

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

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

Answer 1

The correct option is B -1,2

$\left|\begin{array}{ccc}1& 4& 20\\ 1& -2& 5\\ 1& 2x& 5x^2\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}0& 6& 15\\ 0& -2-2x& 5(1-x^2)\\ 1& 2x& 5x^2\end{array}\right|=0\left(\begin{array}{c}{R}_1\to R_1-R_2\\ R_2\to R_2-R_3\end{array}\right)\Rightarrow \mathrm{3.2.5}\left|\begin{array}{ccc}0& 1& 1\\ 0& -(1+x)& 1-x^2\\ 1& x& x^2\end{array}\right|=0\Rightarrow (1+x)\left|\begin{array}{ccc}0& 1& 1\\ 0& -1& 1-x\\ 1& x& x^2\end{array}\right|=0\Rightarrow x+1=0orx-2=0\Rightarrow x=-1,2.$
Trick : Obviously by inspection, x = -1,2 Obviously by inspection.
At $x=-1,\left|\begin{array}{ccc}1& 4& 20\\ 1& -2& 5\\ 1& -2& 5\end{array}\right|=0as{R}_2\equiv R_3$
At $x=2,\left|\begin{array}{ccc}1& 4& 20\\ 1& -2& 5\\ 1& 4& 20\end{array}\right|=0as{R}_1\equiv R_3.$