Question

If $\frac{7}{2}$ and $1$ are the roots of the equation $\left|\begin{array}{ccc}2x& 3& 7\\ 2& 2x& 2\\ 7& 6& 2x\end{array}\right|=0$, then the third root is

• A. $-\frac{7}{2}$
• B. $-\frac{9}{2}$
• C. $-\frac{3}{2}$
• D. $-\frac{5}{2}$

Answer 1

Answer: (B)

$-\frac{9}{2}$

Given
$\left|\begin{array}{ccc}2x& 3& 7\\ 2& 2x& 2\\ 7& 6& 2x\end{array}\right|=0$
$R_1\to R_1+R_2+R_3$
$\Rightarrow \left|\begin{array}{ccc}2x+9& 2x+9& 2x+9\\ 2& 2x& 2\\ 7& 6& 2x\end{array}\right|=0$
$\Rightarrow (2x+9)\left|\begin{array}{ccc}1& 1& 1\\ 2& 2x& 2\\ 7& 6& 2x\end{array}\right|=0$
Given $\frac{7}{2}$ and $1$ are roots of above equation and $(2x+9)$ is a factor of above equation
$\therefore$ $2x+9=0$
i.e., $x=\frac{-9}{2}$ is also root of above equation.