Question

If $(5+2\sqrt{6}{)}^{(x^2-3)}+(5-2\sqrt{6}{)}^{(x^2-3)}=10$, the $x$ is equal to

• A. $\pm 3$ or $\pm \sqrt{3}$
• B. $\pm 5$ or $\sqrt{5}$
• C. $\pm 4$ or $\sqrt{4}$
• D. $\pm 2$ or $\pm \sqrt{2}$

Answer 1

The correct option is D $\pm 2$ or $\pm \sqrt{2}$

Let $(5+2\sqrt{6}{)}^{x^2-3}=a$
We get, $(5-2\sqrt{6}{)}^{x^2-3}=\frac{1}{a}$
$\therefore a+\frac{1}{a}=10\Rightarrow a^2-10a+1=0$
$\Rightarrow a=\frac{10\pm \sqrt{100-4}}{2}=5\pm 2\sqrt{6}$
$=(5+2\sqrt{6}{)}^{(x^2-3)}$ (assumed)
$\Rightarrow x^2-3=\pm 1$
$\Rightarrow x^2=4$ or $2$
$\Rightarrow x^2=4$
$\Rightarrow x=\pm 2$
or $x^2=2$
$\Rightarrow x=\pm 2$ or $\pm \sqrt{2}$.