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

Question

1368 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-12T08:11:27

#x=-10#

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#

Answer 2 · 2017-04-12T08:45:48

#color(blue)(x=22#

#(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