SOLUTION: My homework question is:
Solve (z+2)^-2/3=4
This is what I did:
2^-z/3 + 2^-2/3 = 3
[1/(z+2)^2/3]^3 = 3^3
1/(z+2)^2 = 27
1/2^2+4z+4 = 27
1= 27(z^2+4z+4)
1
Algebra ->
Radicals
-> SOLUTION: My homework question is:
Solve (z+2)^-2/3=4
This is what I did:
2^-z/3 + 2^-2/3 = 3
[1/(z+2)^2/3]^3 = 3^3
1/(z+2)^2 = 27
1/2^2+4z+4 = 27
1= 27(z^2+4z+4)
1
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Question 148978: My homework question is:
Solve (z+2)^-2/3=4
This is what I did:
2^-z/3 + 2^-2/3 = 3
[1/(z+2)^2/3]^3 = 3^3
1/(z+2)^2 = 27
1/2^2+4z+4 = 27
1= 27(z^2+4z+4)
1= 27z^2+108z+108
-1 from each side
27z^2+108z+107=0
From here I think Im supposed to factor but dont know what to do?
Help? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Solve (z+2)^(-2/3)=4
Raise both sides to the (-3/2) power:
[(z+2)^(-2/3)]^(-3/2) = 4^(-3/2)
(z+2) = 2^-3
z+2 = (1/8)
z = (1/8)-(16/8)
z = -15/8
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Cheers,
Stan H.
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