Question

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

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

Answer 1

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

$(5+2\sqrt{6}{)}^{(x^2-3)}+(5-2\sqrt{6}{)}^{(x^2-3)}=10$

$\Rightarrow (5+2\sqrt{6}{)}^{(x^2-3)}+{(5-2\sqrt{6}\times \frac{5+2\sqrt{6}}{5+2\sqrt{6}})}^{(x^2-3)}=10$

$\Rightarrow (5+2\sqrt{6}{)}^{(x^2-3)}+{\left(\frac{1}{5+2\sqrt{6}}\right)}^{(x^2-3)}=10$

Put $t=(5+2\sqrt{6}{)}^{(x^2-3)}$

$\Rightarrow t+\frac{1}{t}=10$
$\Rightarrow t^2-10t+1=0$
$\Rightarrow t=5\pm 2\sqrt{6}$

Case1:
$(5+2\sqrt{6}{)}^{(x^2-3)}=(5+2\sqrt{6})$

$\Rightarrow x^2-3=1\Rightarrow x=\pm 2$

Case2:
$(5+2\sqrt{6}{)}^{(x^2-3)}=(5-2\sqrt{6})$

$\Rightarrow (5+2\sqrt{6}{)}^{(x^2-3)}=(5+2\sqrt{6}{)}^{-1}$

$\Rightarrow x^2-3=-1\Rightarrow x=\pm \sqrt{2}$

Hence, $x=\pm 2$ or $\pm \sqrt{2}$