Question

The value of $x$, which satsfies ${(5+2\sqrt{6})}^{x^2-3}+{(5-2\sqrt{6})}^{x^2-3}=10$ is/are

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

Answer 1

Answer: (A), (D)

The correct options are
A $\pm 2$
D $\pm \sqrt{2}$

Substitute ${(5+2\sqrt{6})}^{x^2-3}=t$
$\therefore {\left(\frac{(5+2\sqrt{6})(5-2\sqrt{6})}{(5-2\sqrt{6})}\right)}^{x^2-3}=t$
${\left(\frac{1}{(5-2\sqrt{6})}\right)}^{x^2-3}=t$
$\therefore {(5-2\sqrt{6})}^{x^2-3}=\frac{1}{t}$
Then given equation becomes
$t+\frac{1}{t}=10$
$t^2-10t+1=0$
$t=\frac{10\pm \sqrt{100-4}}{2}$
$t=5\pm 2\sqrt{6}$
Thus, ${(5+2\sqrt{6})}^{x^2-3}={(5+2\sqrt{6})}^{\pm 1}$
$x^2-3=\pm 1$
$\therefore x^2-3=1or{x}^2-3=-1$
$x^2=4or{x}^2=2$
$x=\pm 2,\pm \sqrt{2}$