How do you solve #(x-6)^(1/2) +4=0# and find any extraneous solutions?

2 Answers
Apr 12, 2017

Answer:

#x=-10#

Explanation:

Get rid of square root first:
#((x-6)^0.5)^2+4^2=0#
#(x-6)=-16#
#x=-16+6=-10#

This is the solution of your equation. #x=-10#

Apr 12, 2017

Answer:

#color(blue)(x=22#

Explanation:

#(x-6)^(1/2)+4=0#

#:.(x-6)^(1/2)=-4#

#:.sqrt(x-6)=-4#

square both sides

#:.(sqrt(x-6))^2=(-4)^2#

#:.sqrt(x-6) xx sqrt(x-6)=x-6#

#:.x-6=16#

#:.x=16+6#

#:.color(blue)(x=22#

sustitute #x=22#

#:.sqrt(22-6)+4=0#

#:.+-sqrt(16)+4=0#

#:.+-4+4=0#

#:.color(blue)(-4+4=0#
#4+4=# extraneous solution