How do you solve #sqrt(r+3)=r-3# and check your solution?

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Answer 1 · 2017-06-06T14:48:44

#r=6#

#sqrt(r+3)=r-3#

#r+3=(r-3)^2#

#r+3=r^2-6r+9#

#r^2-7r+6=0#

#(r-1)(r-6)=0#

#r=6# or #r=1#

#sqrt(6+3)-=6-3# so #r=6 # is a valid solution

#sqrt(1+3)!=1-3# so #r=1# is not a valid solution