[tex] \sqrt{2z - 6} - \sqrt{z + 6} = 0 [/tex]
[tex] \sqrt{2z - 6} = \sqrt{z + 6} [/tex]
[tex] (\sqrt{2z - 6})^2 = (\sqrt{z + 6})^2 [/tex]
[tex] 2z - 6 = z + 6 [/tex]
[tex] z - 6 = 6 [/tex]
[tex] z = 12 [/tex]
Since we squared both sides, we must check the solution on the original equation.
[tex] \sqrt{2(12) - 6} - \sqrt{12 + 6} = 0 [/tex]
[tex] \sqrt{24 - 6} - \sqrt{18} = 0 [/tex]
[tex] \sqrt{18} - \sqrt{18} = 0 [/tex]
[tex] 0 = 0 [/tex]
Since the solution makes for a true statement, z = 12 is the solution.
Answer: D. z = 12
