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

1 Answer
Aug 4, 2015

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

Explanation:

Given -z = sqrt(-z+6)z=z+6
Square both sides
color(white)("XXXX")XXXXz^2 = -z+6z2=z+6
Re-arrange into standard form
color(white)("XXXX")XXXXz^2+z-6=0z2+z6=0
Factor
color(white)("XXXX")XXXX(z+3)(z-2)=0(z+3)(z2)=0

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

Checking
color(white)("XXXX")XXXXwith z=-3z=3
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXX-z = sqrt(-z+6)z=z+6
color(white)("XXXX")XXXXbecomes
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXX3 = sqrt(3+6) = 33=3+6=3
color(white)("XXXX")XXXXValid

color(white)("XXXX")XXXXwith z=2z=2
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXX-z=sqrt(-z+6)z=z+6
color(white)("XXXX")XXXXbecomes
color(white)("XXXX")XXXXcolor(white)("XXXX")XXXX-2 ?= sqrt(-2+6) = 22?=2+6=2
color(white)("XXXX")XXXXExtraneous