Question

Let $D _1=\begin{vmatrix}x & a & b \\ -1 & 0 & x \\ x & 2 & 1\end{vmatrix}$ and $D _2=\begin{vmatrix}cx ^2 & 2 a & - b \\ x & 2 & 1 \\ -1 & 0 & x \end{vmatrix}$.
If all the roots of the equation $\left(x^2-4 x-7\right)\left(x^2-2 x-3\right)=0$ satisfies the equation $D_1+D_2=0$ then find the value of $(a+4 b+c)$.

Answer 1

$D_1+D_2=0$
$\begin{vmatrix}x & a & b \\ -1 & 0 & x \\ x & 2 & 1\end{vmatrix}-\begin{vmatrix}x ^2 & 2 a & - b \\ -1 & 0 & x \\ x & 2 & 1\end{vmatrix}=0$
$\begin{vmatrix}x-c x^2 & -a & 2 b \\ -1 & 0 & x \\ x & 2 & 1\end{vmatrix}=0$
$\left(x-c x^2\right)(-2 x)+a\left(-1-x^2\right)+2 b(-2)=0 $
$-2 x^2+2 c x^3-a-a x^2-4 b=0$
$2 c x^3-(a+2) x^2-(a+4 b)=0$
The above equation is satisfied by four different values of $x$, $\therefore$ It is an identity.
$c=0, a+2=0 \Rightarrow a=-2, a+4 b=0 \Rightarrow b=\frac{1}{2} $
$\therefore a+4 b+c=0$