A) \[3\pm 2\sqrt{2}\]
B) \[\pm 1\]
C) \[\pm 3\sqrt{3},\pm 2\sqrt{2}\]
D) \[\pm 7,\pm \sqrt{3}\]
E) \[\pm 3,\pm \sqrt{7}\]
Correct Answer: E
Solution :
Given, \[{{(3+2\sqrt{2})}^{{{x}^{2}}-8}}+{{(3+2\sqrt{2})}^{8-{{x}^{2}}}}=6\] Let \[{{(3+2\sqrt{2})}^{{{x}^{2}}-8}}=y\] \[\therefore \] \[y+{{y}^{-1}}=6\] \[\Rightarrow \] \[{{y}^{2}}-6y+1=0\] \[\Rightarrow \] \[y=\frac{6\pm \sqrt{36-4}}{2\times 1}\] \[=\frac{6\pm 4\sqrt{2}}{2}=3\pm 2\sqrt{2}\] For\[+\]sign \[{{(3+2\sqrt{2})}^{{{x}^{2}}-8}}=3+2\sqrt{2}\] \[\Rightarrow \] \[{{x}^{2}}-8=1\Rightarrow x=\pm 3\] For\[-\]sign \[{{[{{(3+2\sqrt{2})}^{-1}}]}^{8-{{x}^{2}}}}=3-2\sqrt{2}\] \[\Rightarrow \] \[{{(3-2\sqrt{2})}^{8-{{x}^{2}}}}=3-2\sqrt{2}\] \[\Rightarrow \] \[8-{{x}^{2}}=1\Rightarrow {{x}^{2}}=7\] \[\Rightarrow \] \[x=\pm \sqrt{7}\]
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