Question

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

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

Answer 1

Answer: (D)

$2$

Let $\sqrt{a\sqrt{a\sqrt{a................\infty }}}=k$

$\sqrt{ak}=k$

$k=a$

so, $\sqrt{a\sqrt{a\sqrt{a................\infty }}}=a$ ............1

similarly,Let $\sqrt{x\sqrt{x\sqrt{x................\infty }}}=j$

$\sqrt{xj}=j$

$j=x$

so, $\sqrt{x\sqrt{x\sqrt{x................\infty }}}=x$ ...............2

And , $\sqrt{49+20\sqrt{6}}=\sqrt{25+24+20\sqrt{6}}=\sqrt{(5+2\sqrt{6}{)}^2}=5+2\sqrt{6}$ ................3

Given , $a=x^2-3$

$({\sqrt{49+20\sqrt{6}})}^{\sqrt{a\sqrt{a\sqrt{a.......\infty }}}}+{(5-2\sqrt{6})}^{x^2+x-3-\sqrt{x\sqrt{x\sqrt{x..........\infty }}}}=10$

By using 1,2&3

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

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

Hence , $x^2-3=1$

$x^2=4$

$x=2,-2$

and x must be positive

so $x=2$