How do you solve a problem like this by taking the square of boyh sides?
x+2>square root of x^2-16
Answer is : x>=4.Square root can't be-.But there is not a method to solve this one.
first check domain on both sides for sqrt(x^2-16) it is (-infinity to -4]U[4,infinity)
and for x+2 it is any real no
now sqrt(x^2-16)>=0
so x+2 takes positive values
so it can have domain as [0,infinity)
but we need to take in consideration the right hand side which contains square root term
if we split [0,infinity) into two parts i.e.
[0,4)U[4,infinity)
we see that [0,4) does not satisfy the domain for right hand side
so x belongs to [4,infinity)