Question

If ${\left(\sqrt{(49+20\sqrt{6})}\right)}^{\sqrt{a\sqrt{a\sqrt{a....\infty }}}}+(5-2\sqrt{6}{)}^{x^2+x-3-\sqrt{x\sqrt{x\sqrt{x....\infty }}}}=10$, where $a=x^2-3$, then $x$ is equal to

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

Answer 1

The correct option is D $2$

${\left(\sqrt{(49+20\sqrt{6})}\right)}^{\sqrt{a\sqrt{a\sqrt{a....\infty }}}}+(5-2\sqrt{6}{)}^{x^2+x-3-\sqrt{x\sqrt{x\sqrt{x....\infty }}}}=10$ where $a=x^2-3$
$\sqrt{a\sqrt{a\sqrt{a....\infty }}}=a=x^2-3$
$\Rightarrow (5+2\sqrt{6}{)}^{x^2-3}+(5-2\sqrt{6}{)}^{x^2+x-3-x}=10$
Where $a=x^2-3>0$ and $x>0$
$\Rightarrow (5+2\sqrt{6}{)}^{x^2-3}+(5-2\sqrt{6}{)}^{x^2-3}=10$
Let $(5+2\sqrt{6}{)}^{x^2-3}=t$
$\Rightarrow t^2-10t+1=0$
$\Rightarrow t=5\pm 2\sqrt{6}$
$\Rightarrow x^2-3=\pm 1$
$\Rightarrow x=\pm \sqrt{2}$ or $x=\pm 2$
But $a=x^2-3>0$ and $x>0$
$\therefore x=2$
Hence, option D .