question_answer

Show that x = 2 is a root of the equation formed by the following determinant. Hence, solve the equation.

Answer:

Let Expand along \[{{R}_{1}},\]we get \[|A|\,\,=x\,(+-\,3{{x}^{2}}-6x-2{{x}^{2}}+6x)\] \[+\,\,6\,(2x+4+3x-9)-1\,(4x-9x)\] \[=x\,(-\,5{{x}^{2}})+6\,(5x-5)-1\,(-\,5x)\] \[=-\,5{{x}^{3}}+30x-30+5x\] \[=-\,5{{x}^{3}}+35x-30\] For \[x=2\] be root of \[|A|,\] \[|A|\,\,=0\] i,e         \[|A|\,\,=-\,5\,{{(2)}^{3}}+35\,(2)-30\] \[=-\,5\,(8)+70-30\] \[=-\,40+70-30\] \[=-\,70+70=0\]             Hence, \[x=2\] is a root of the equation. Now, as \[(x-2)\] is a root of the equation, then \[-\,{{5}^{2}}(x-2)-10x\,(x-2)+15\,(x-2)=0\] \[\Rightarrow \]   \[(x-2)\,(-\,5{{x}^{2}}-10x+15)\] \[\Rightarrow \]   \[-\,5\,(x-2)\,\,({{x}^{2}}+2x-3)=0\] For other roots of the equation, put \[{{x}^{2}}+2x-3=0\Rightarrow {{x}^{2}}+3x-x-3=0\] \[\Rightarrow \]\[x\,(x+3)-1\,(x+3)=0\Rightarrow (x+3)\,\,(x-1)=0\] \[\Rightarrow \]\[x+3=0\]and \[x-1=0\Rightarrow x=-\,3\]and \[x=1\] Thus, \[x=-\,3\]and 1.