Question

Let $D_1=\left|\begin{array}{ccc}x& a& b\\ -1& 0& x\\ x& 2& 1\end{array}\right|$ and $D_2=\left|\begin{array}{ccc}c{x}^2& 2a& -b\\ x& 2& 1\\ -1& 0& x\end{array}\right|.$ If all the roots of the equation $(x^2-4x-7)(x^2-2x-3)=0$ satisfy the equation $D_1+D_2=0,$ then the value of $a+4b+c$ is

Answer 1

$D_1+D_2=0\Rightarrow \left|\begin{array}{ccc}x& a& b\\ -1& 0& x\\ x& 2& 1\end{array}\right|+\left|\begin{array}{ccc}c{x}^2& 2a& -b\\ x& 2& 1\\ -1& 0& x\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x& a& b\\ -1& 0& x\\ x& 2& 1\end{array}\right|-\left|\begin{array}{ccc}c{x}^2& 2a& -b\\ -1& 0& x\\ x& 2& 1\end{array}\right|=0$ [Row interchange]

$\Rightarrow \left|\begin{array}{ccc}x-c{x}^2& -a& 2b\\ -1& 0& x\\ x& 2& 1\end{array}\right|=0$

$\Rightarrow (x-c{x}^2)(-2x)+a(-1-x^2)+2b(-2)=0$
$\Rightarrow -2x^2+2c{x}^3-a-a{x}^2-4b=0$
$\Rightarrow 2c{x}^3-(a+2)x^2-(a+4b)=0$
The above equation is satisfied by four diferent values of $x.$
$\therefore$ It is an identity in $x.$
So, $c=0$
$a+2=0\Rightarrow a=-2$
and $a+4b=0\Rightarrow b=\frac{1}{2}$
$\therefore a+4b+c=0$