Question

The solution of the equation ${(3+2\sqrt{2})}^{{\mathrm{x}}^2-8}+{(3+2\sqrt{2})}^{8-{\mathrm{x}}^2}=6$ are

• 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. $\pm 3,\pm \sqrt{7}$

Answer 1

The correct option is E

$\pm 3,\pm \sqrt{7}$

The explanation for the correct answer.

Find the solution to the given equation:

Given: ${(3+2\sqrt{2})}^{{\mathrm{x}}^2-8}+{(3+2\sqrt{2})}^{8-{\mathrm{x}}^2}=6$

Let ${(3+2\sqrt{2})}^{{\mathrm{x}}^2-8}=\mathrm{t}$

$$\begin{gathered} \Rightarrow \mathrm{t}+\frac{1}{\mathrm{t}}=6 \\ \Rightarrow \frac{{\mathrm{t}}^2+1}{\mathrm{t}}=6 \\ \Rightarrow {\mathrm{t}}^2+1=6\mathrm{t} \\ \Rightarrow {\mathrm{t}}^2-6\mathrm{t}+1=0 \\ \Rightarrow \mathrm{t}=\frac{6\pm \sqrt{36-4}}{2}\left[\mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}\right] \\ \Rightarrow \mathrm{t}=\frac{6\pm \sqrt{32}}{2} \\ \Rightarrow \mathrm{t}=\frac{6\pm 4\sqrt{2}}{2} \\ \Rightarrow \mathrm{t}=3\pm 2\sqrt{2} \end{gathered}$$

Now compare the values

$$\begin{gathered} \Rightarrow {(3+2\sqrt{2})}^{{\mathrm{x}}^2-8}=3+2\sqrt{2}\mathrm{or}{(3+2\sqrt{2})}^{8-{\mathrm{x}}^2}=3-2\sqrt{2} \\ \Rightarrow {\mathrm{x}}^2-8=1\mathrm{or}{(3+2\sqrt{2})}^{8-{\mathrm{x}}^2}=3-2\sqrt{2}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}} \\ \Rightarrow {\mathrm{x}}^2=9\mathrm{or}{(3+2\sqrt{2})}^{8-{\mathrm{x}}^2}=\frac{3^2-2{\sqrt{2}}^2}{3+2\sqrt{2}} \\ \Rightarrow \mathrm{x}=\pm 3\mathrm{or}{(3+2\sqrt{2})}^{8-{\mathrm{x}}^2}=\frac{9-8}{3+2\sqrt{2}} \\ \Rightarrow \mathrm{or}{(3+2\sqrt{2})}^{8-{\mathrm{x}}^2}=\frac{1}{3+2\sqrt{2}} \\ \Rightarrow \mathrm{or}{(3+2\sqrt{2})}^{8-{\mathrm{x}}^2}={\left(3+2\sqrt{2}\right)}^{-1} \\ \Rightarrow \mathrm{or}8-{\mathrm{x}}^2=-1 \\ \Rightarrow \mathrm{or}{\mathrm{x}}^2=7 \\ \Rightarrow \mathrm{or}\mathrm{x}=\pm \sqrt{7} \end{gathered}$$

Hence option $E$ is the correct answer.