A) \[(-4,5)\]
B) \[(4,5)\]
C) \[(-4,-5)\]
D) \[(-4,3)\]
Correct Answer: A
Solution :
\[\left| \begin{matrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \\ \end{matrix} \right|\] \[{{\text{R}}_{\text{1}}}\to {{\text{R}}_{\text{1}}}\text{+}{{\text{R}}_{\text{2}}}\text{+}{{\text{R}}_{\text{3}}}\] \[\left| \begin{matrix} 5x-4 & 5x-4 & 5x-4 \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \\ \end{matrix} \right|\] \[(5x-4)\left| \begin{matrix} 1 & 1 & 1 \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \\ \end{matrix} \right|\] \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\] \[(5x-4)\left| \begin{matrix} 0 & 0 & 0 \\ x+4 & -x-4 & 2x \\ 0 & x+4 & x-4 \\ \end{matrix} \right|\] \[={{(x+4)}^{2}}(5x-4)=(A+Bx){{(x-A)}^{2}}\] \[A=-4,B=5\Rightarrow (-4,5)\]
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