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 $\left(x^2-4x-7\right)\left(x^2-2x-3\right)=0$ satisfies the equation $D_1+D_2=0$ then find the value of $(a+4b+c)$.

Answer 1

Answer: 0

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