A) 3, 3.5
B) 1, 3.5
C) 1, 7
D) 2, 7
Correct Answer: B
Solution :
\[\left| \begin{matrix} 2x+1 & 4 & 8 \\ 2 & 2x & 2 \\ 7 & 6 & 2x \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix} 2x+1 & 2x+10 & 2x+10 \\ 2 & 2x & 2 \\ 7 & 6 & 2x \\ \end{matrix} \right|=0\] \[({{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}})\] \[\Rightarrow \] \[(2x+10)\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 2x & 2 \\ 7 & 6 & 2x \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(2x+10)\left| \begin{matrix} 1 & 0 & 0 \\ 2 & 2x-2 & 0 \\ 7 & -1 & 2x-7 \\ \end{matrix} \right|=0\] (\[{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}\]and \[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}}\]) \[\Rightarrow \]\[(2x+10)(2x-2)(2x+7)=0\] \[\Rightarrow \] \[x=-5,1,\frac{7}{2}\] Hence, other roots are 1 and \[\frac{7}{2}\].
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