Question #26706

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Answer 1 · 2017-11-12T01:21:39

change back into radical form, and solve.

#sqrt(x^2-1) = 0 #

we can square both sides because of the square root property

#(sqrt(x^2-1))^2 = (0)^2 #

#x^2-1=0#

#x^2-1# is a difference of squares

#(x-1)(x+1) = 0#

we take each factor and equal it to #0#

#(x-1 )= 0#
#x = 1#

#(x+1) = 0#
#x = -1#

so our domain is #x<= -1# and #x>=1# which is the set of #RR#

to be clear #0# is not included because Domain corresponds to the X values where y is #0#

Side note if 0 was allowed, #sqrt(0^2-1)# will be #sqrt(-1)# which is an imaginary value, and not in the set of #RR#

set builder #{x in RR | x<=-1 or x>=1}#