What is #(sqrt(2x+4))^2#? Algebra 1 Answer George C. Aug 9, 2015 #(sqrt(2x+4))^2 = 2x+4# for all #x in RR# or for all #x in [-2, oo)# if you only consider #sqrt# as a real valued function. Explanation: Note that if #x < -2# then #2x+4 < 0# and #sqrt(2x+4)# has a complex (pure imaginary) value, but its square will still be #2x+4#. Essentially, #(sqrt(z))^2 = z# by definition. If the square root exists, then it is a value whose square gives you back the original value. Interestingly, #sqrt((2x+4)^2) = abs(2x+4)# not #2x+4# Answer link Related questions How do I determine the molecular shape of a molecule? What is the lewis structure for co2? What is the lewis structure for hcn? How is vsepr used to classify molecules? What are the units used for the ideal gas law? How does Charle's law relate to breathing? What is the ideal gas law constant? How do you calculate the ideal gas law constant? How do you find density in the ideal gas law? Does ideal gas law apply to liquids? Impact of this question 1364 views around the world You can reuse this answer Creative Commons License