What is #(sqrt(2x+4))^2#?

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#

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