Nested Radicals

I assume you're referring to the challenge master note for Akagami Ng. He defined the expression in question to be of value of $x$ . It can be easily shown that they are two positive numbers, so the multiplication of two positive numbers is definitely positive, so $x> 0$ . In Akagami Ng's solution, he square both sides of the equation in the very second step. When you raise both sides to the power of some integer other than 1 or -1, you might get an extraneous root. For a simpler example, if $x=1$ , square both sides gives $x^2 =1$ which means that $x^2-1 = 0$ or $(x-1)(x+1) = 0$ or $x = 1, -1$ . This shows that there's one "extra" root of $-1$ , because if $x=-1$ and it is given that $x=1$ then $-1=1$ which doesn't make sense, so $-1$ is called an "extraneous root". We need to be careful to check back whether a value is actually a solution (that fits the criteria), in this case $x>0$ so $x\ne -2$ . Hope this clears things up.