Question

If one of the root of the equation $\left|\begin{array}{ccc}7& 6& x\\ 2& x& 2\\ x& 3& 7\end{array}\right|=0$ is $-9$, then other roots are

• A. $2,6$
• B. $3,6$
• C. $2,7$
• D. $3,7$

Answer 1

Answer: (C)

$2,7$

Given, $\left|\begin{array}{ccc}7& 6& x\\ 2& x& 2\\ x& 3& 7\end{array}\right|=0$

$\Rightarrow 7(7x-6)-6(14-2x)+x(6-x^2)=0$

$\Rightarrow 49x-42-84+12x+6x-x^3=0$

$\Rightarrow x^3-67x+126=0$

Since, it is given that $(x+9)$ is a root of the equation $x^3-67x+126=0$. Therefore, $x^3-67x+126=0$ is divisible by $x+9$.

$x^3-67x+126=0$ is divisible by $x+9$.

$\therefore x^3-67x+126=(x+9)(x^2-9x+14)$

$=(x+9)(x-7)(x-2)$

Hence, other roots are $2$ and $7$.