Answer:
w = -11
Step-by-step explanation:
[tex]\sqrt{3 - 2w} = w + 6[/tex]
[tex](\sqrt{3 - 2w})^2 = (w + 6)^2[/tex]
[tex] 3 - 2w = w^2 + 12w + 36 [/tex]
[tex] w^2 + 14w + 33 = 0 [/tex]
[tex] (w + 11)(w + 3) = 0 [/tex]
[tex] w + 11 = 0 [/tex] or [tex] w + 3 = 0 [/tex]
[tex] w = -11 [/tex] or [tex] w = -3 [/tex]
When you square both sides of an equation, you must check all solutions for extraneous solutions.
Check w = -11.
[tex]\sqrt{3 - 2w} = w + 6[/tex]
[tex] \sqrt{3 - 2(-11)} = -11 + 6 [/tex]
[tex] \sqrt{3 + 22} = -5 [/tex]
[tex] \sqrt{25} = -5 [/tex]
[tex] 5 = -5 [/tex]
This is a false statement, so the solution w = -11 is extraneous since it does not satisfy the original equation.
Check w = -3.
[tex]\sqrt{3 - 2w} = w + 6[/tex]
[tex] \sqrt{3 - 2(-3)} = -3 + 6 [/tex]
[tex] \sqrt{3 + 6} = 3 [/tex]
[tex] \sqrt{9} = 3 [/tex]
[tex] 3 = 3 [/tex]
This is a true statement, so the solution w = -3 is valid.
Answer: w = -11
