There seem to me to be two aspects to this question:

(1) What does "square root of #x^2+4#" mean?

#sqrt(x^2+4)# is a term which when squared yields #x^2+4# :

#sqrt(x^2+4) xx sqrt(x^2+4) = x^2 + 4#

In other words #t = sqrt(x^2+4)# is the solution #t# of the

equation #t^2 = x^2+4#

(2) Can the formula #sqrt(x^2+4)# be simplified?

No.

For starters #(x^2+4) > 0# for all #x in RR#, so it has no linear factors with real coefficients.

Suppose you produced some formula #f(x)# for #sqrt(x^2+4)#. Then #f(1) = sqrt(5)# and #f(2) = sqrt(8) = 2 sqrt(2)#.

So any such formula #f(x)# would involve square roots or fractional exponents or suchlike, and be as complex as the original #sqrt(x^2+4)#