Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 25

$z=2$ is extraneous

$Given,$ $-z=\sqrt (-z+6)$ Checking for $z=-3$ : $L.H.S=-(-3)=3$ $R.H.S=\sqrt (-(-3)+6)=\sqrt 9=3$ $L.H.S=R.H.S$ , Thus not extraneous Checking for $z=2$ $L.H.S=-2$ $R.H.S=\sqrt (-2+6)=\sqrt 4=2$ $L.H.S$ does not equal $R.H.S$,thus $z=2$ is extraneous