Question 852097
{{{x^4-16=0}}}
{{{x^4=16}}}
{{{sqrt(x^4)=sqrt(16)}}}
{{{x^2=4}}}
{{{sqrt(x^2)=sqrt(4)}}}
{{{x=2}}}

But we also have to consider the negative and imaginary solutions as well. 
The former gives us 
   X=2, -2
For the latter {{{i^2=-1}}} and hence {{{i^4=(i^2)^2=1}}} . 
This gives us the following two additional solutions
   X=2i, -2i