Question

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

• A. $-\sqrt{2},\sqrt{2}$
• B. $-2,2$
• C. $2,\sqrt{2}$
• D. $-2,-\sqrt{2}$

Answer 1

Answer: (A), (B)

The correct options are
A $-\sqrt{2},\sqrt{2}$
B $-2,2$

Given,
$(5+2\sqrt{6}{)}^{x^2-3}=10$
we have,
In, $(a+b{)}^{x^2-k}(a-b{)}^{x^2-k}=2a$
if $a^2-b=1$ then $x^2-k=\pm 1$
Here,
$a=5$ and $\sqrt{b}=\sqrt{24}$
$\therefore a^2-b=25-24=1$Hence, $x^2-3=\pm 1$
$x^2-3=1$
$x^2=4$
$x=\pm 2$
or
$x^2-3=-1$
$x^2=2$
$x=\pm \sqrt{2}$

$\therefore x=\pm 2,\pm \sqrt{2}$