How do you solve and check for extraneous solutions in #-z=sqrt( -z+6)#?

1 Answer
Alan P.
Aug 4, 2015

Valid solution #z=-3#
(plus extraneous solution #z=2#)

Explanation:

Given #-z = sqrt(-z+6)#
Square both sides
#color(white)("XXXX")##z^2 = -z+6#
Re-arrange into standard form
#color(white)("XXXX")##z^2+z-6=0#
Factor
#color(white)("XXXX")##(z+3)(z-2)=0#

#rArr#
#color(white)("XXXX")##z = -3 or z = +2#

Checking
#color(white)("XXXX")#with #z=-3#
#color(white)("XXXX")##color(white)("XXXX")##-z = sqrt(-z+6)#
#color(white)("XXXX")#becomes
#color(white)("XXXX")##color(white)("XXXX")##3 = sqrt(3+6) = 3#
#color(white)("XXXX")#Valid

#color(white)("XXXX")#with #z=2#
#color(white)("XXXX")##color(white)("XXXX")##-z=sqrt(-z+6)#
#color(white)("XXXX")#becomes
#color(white)("XXXX")##color(white)("XXXX")##-2 ?= sqrt(-2+6) = 2#
#color(white)("XXXX")#Extraneous

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