Question

6 solve radical equations (homework)
pre: 8/16 answered: 8/16
question 9
find all real solutions:
\\(\sqrt{z + 6}=z + 6\\)
\\(z =\\) . (if no solution exists, enter dne; if more than one solution exists, enter the solutions separated by commas)

Explanation:

Step1: Square both sides

$(\sqrt{z + 6})^2=(z + 6)^2$
$z + 6=z^{2}+12z + 36$

Step2: Rearrange to quadratic form

$z^{2}+12z + 36-(z + 6)=0$
$z^{2}+12z + 36 - z-6=0$
$z^{2}+11z + 30=0$

Step3: Factor the quadratic

$(z + 5)(z+6)=0$

Step4: Solve for z

$z+5 = 0$ gives $z=-5$; $z + 6=0$ gives $z=-6$

Step5: Check solutions

For $z=-5$, $\sqrt{-5 + 6}=-5 + 6$, $\sqrt{1}=1$, $1 = 1$ (valid)
For $z=-6$, $\sqrt{-6 + 6}=-6 + 6$, $\sqrt{0}=0$, $0 = 0$ (valid)

Answer:

$-5,-6$