Number of real values of $x$ : $(5+2 \sqrt{6})^{x^{2}-3}+(5-2 \sqrt{6})^{x^{2}-3}=10$

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Professor PC Thomas Classes · Answer

Let $(5+2 \sqrt{6})^{x^{2}-3}=t$\ then $(5-2 \sqrt{6})^{x^{2}-3}$\ $=\left[\frac{(5-2 \sqrt{6})(5+2 \sqrt{6})}{5+2 \sqrt{6}}ight]^{x^{2}-3}=\left(\frac{1}{t}ight)$\ $	herefore t+\frac{1}{t}=10 \Rightarrow t^{2}-10 t+1=0$\ $	herefore t=\frac{10 \pm \sqrt{100-4}}{2}=\frac{10 \pm 4 \sqrt{6}}{2}$\ $	herefore t=5+2 \sqrt{6}$ or $t=5-2 \sqrt{6}$.\ Consider, $t=5+2 \sqrt{6}$\ $(5+2 \sqrt{6})^{x^{2}-3}=5+2 \sqrt{6}$\ $	herefore \quad x^{2}-3=1 \Rightarrow x^{2}=4 	herefore x= \pm 2$\ Again consider $t=5-2 \sqrt{6}$\ $=\frac{1}{5+2 \sqrt{6}}=(5+2 \sqrt{6})^{-1}$\ $	herefore \quad(5+2 \sqrt{6})^{x^{2}-3}=(5+2 \sqrt{6})^{-1}$\ $	herefore \quad x^{2}-3=-1 	herefore x= \pm \sqrt{2}$\ Hence the solutions are $2,-2, \sqrt{2},-\sqrt{2}$\