An algebra problem by Renz Mina

Let $\sqrt{x \sqrt{x \sqrt{x \ldots}}} = a$ and $\sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} = b.$

Now $a^2 = x \times a \Rightarrow a=x.$

Now, $b^2=x+b \Rightarrow b = \dfrac{1+\sqrt{1+4x}}{2}.$

$\sqrt{x \sqrt{x \sqrt{x \ldots}}} + \sqrt{x + \sqrt{x + \sqrt{x + \ldots}}} = a+b =9.$

Therefore ,

$a+b = x+\dfrac{1+\sqrt{1+4x}}{2} = 9 \Rightarrow x=6.$

Hope everyone likes my solution.

Let $X=\sqrt{x\cdot \sqrt{x \cdot \sqrt{x \cdot ... }}}$ and $Y=\sqrt{x+ \sqrt{x + \sqrt{x + ... }}}$ . We have, in the first place, that

$X^2=x \cdot X \implies X=x$

and, in the second place, that

$Y^2=x+Y \implies Y=\displaystyle\frac{1+\sqrt{1+4x}}{2}$

So, we can conclude that

$\sqrt{x\cdot \sqrt{x \cdot \sqrt{x \cdot ... }}}+\sqrt{x+ \sqrt{x + \sqrt{x + ... }}}-9=0 \implies$ $x+\displaystyle\frac{1+\sqrt{1+4x}}{2}=9 \implies x=6$

$\sqrt{x{ \sqrt{x{\sqrt{x ...}}}}}+\sqrt{x{+ \sqrt{x+\sqrt{x +...}}}}-9=0$ $x^{ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+...}+\sqrt{x{+ \sqrt{x +\sqrt{x +...}}}}-9=0$

$x^ \frac{{ \frac{1}{2}}}{{1-\frac{1}{2}}}+\sqrt{x{+ \sqrt{x +\sqrt{x +...}}}}-9=0$ $\sqrt{x{+ \sqrt{x +\sqrt{x +...}}}}=9-x (*)$

Square $(*)$ it, $x{+ \sqrt{x+\sqrt{x +\sqrt{x +...}}}}={(9-x )}^2(**)$ (with $x<9 ).$ Subtitution $(*)$ to $(**)$ we can have $x+(9-x)={(9-x)}^2 \implies x=6$ (because $x=12>9).$

$x= \boxed6$