How does anyone solve this equation on $x$ ($x$ is a real number)?
$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}=x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{\cdots}}}$$
Thank you in advance.
How does anyone solve this equation on $x$ ($x$ is a real number)?
$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}=x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{\cdots}}}$$
Thank you in advance.
$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}=y;\quad y=x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{\cdots}}}$$
System $\sqrt{x+y}=y;\quad y=x+\dfrac{x}{y}$
both $x$ and $y$ are positive
$x+y=y^2\to x=y^2-y$
plug in the second equation
$y=y^2-y+\dfrac{y^2-y}{y}\to y=y^2-y+y-1$
$y^2-y-1=0\to y=\dfrac{1+\sqrt{5}}{2}$
and $x=\left(\dfrac{1+\sqrt{5}}{2}\right)^2-\dfrac{1+\sqrt{5}}{2}$
$x=1$
Hope it helps
Note that if $L(x)$ and $R(x)$ are the left-hand side and the right-hand side respectively, then $$L(x)=\sqrt{x+L(x)}=x+\frac{x}{R(x)}=R(x).$$ Hence $L(x)> 0$, $R(x)>0$, $$L^2(x)=L(x)+x\quad\mbox{and}\quad R^2(x)=xR(x)+x.$$ Finally, from $L(x)=R(x)$, we obtain $$L(x)+x=xL(x)+x\implies (1-x)L(x)=0\implies x=1.$$
P.S. Note that $$L(x)=\frac{1+\sqrt{1+4x}}{2}\quad\mbox{and}\quad R(x)=\frac{x+\sqrt{x^2+4x}}{2}.$$