How do you solve #x+2-2sqrt(x+3)=0# and find any extraneous solutions?

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Answer 1 · 2017-10-27T16:03:08

#x+2-2sqrt(x+3)=0#

#x+2=2sqrt(x+3)#

Raise to the square. Remember that this will add false solutions:

#(x+2)^2=(2sqrt(x+3))^2#

#x^2+4x+4=4(x+3)#

#x^2+4x+4=4x+12#

#x^2+4=+12#
#x^2-8=0#
#x=+-sqrt(8)#

Now verify the solutions for looking for false solutions:

only #+sqrt(8)# is correct.