Question

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

Answer 1

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