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

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

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Answer 1 · 2015-08-04T18:10:47

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

Explanation:

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

$\Rightarrow$ $rArr$
$XXXX$ $color(white)("XXXX")$ $z = - 3 or z = + 2$ $z = -3 or z = +2$

Checking
$XXXX$ $color(white)("XXXX")$ with $z = - 3$ $z=-3$
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $- z = \sqrt{- z + 6}$ $-z = sqrt(-z+6)$
$XXXX$ $color(white)("XXXX")$ becomes
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $3 = \sqrt{3 + 6} = 3$ $3 = sqrt(3+6) = 3$
$XXXX$ $color(white)("XXXX")$ Valid

$XXXX$ $color(white)("XXXX")$ with $z = 2$ $z=2$
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $- z = \sqrt{- z + 6}$ $-z=sqrt(-z+6)$
$XXXX$ $color(white)("XXXX")$ becomes
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $- 2 ? = \sqrt{- 2 + 6} = 2$ $-2 ?= sqrt(-2+6) = 2$
$XXXX$ $color(white)("XXXX")$ Extraneous

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

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Explanation:

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