SOLUTION: (v^2 - 4v)^2 - 17(v^2 - 4v) +60 = 0 This is what I have so far? Not sure where I went wrong. let v = v^2 - 4v v^2 - 17v +60 = 0 (v-12)(v-5) = 0 v-12 = 0 or v-5 = 0 v=12

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Question 1016841: (v^2 - 4v)^2 - 17(v^2 - 4v) +60 = 0
This is what I have so far? Not sure where I went wrong.
let v = v^2 - 4v
v^2 - 17v +60 = 0
(v-12)(v-5) = 0
v-12 = 0 or v-5 = 0
v=12 or v=5
v^2 = 12 or v^2 = 5
√v^2 = =-√12 or √v^2 = +-√5
v = +-2√3 or v = +-√5
Found 2 solutions by richard1234, MathTherapy:
Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
You should use a different letter than "v" for your substitutions since it is already a variable name (say "u = v^2 - 4v").

The mistake comes where you say "u=12 or u=5" and the next line is "v^2 = 12 or v^2 = 5." Should be "v^2 - 4v = 12 or "v^2 - 4v = 5."
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

(v^2 - 4v)^2 - 17(v^2 - 4v) +60 = 0
This is what I have so far? Not sure where I went wrong.
let v = v^2 - 4v
v^2 - 17v +60 = 0
(v-12)(v-5) = 0
v-12 = 0 or v-5 = 0
v=12 or v=5
v^2 = 12 or v^2 = 5
√v^2 = =-√12 or √v^2 = +-√5
v = +-2√3 or v = +-√5
The first mistake you made was to substitute the same variable for itself. This can certainly can be very confusing.

Therefore, I'd suggest that you not do so.



Let 

 ------- Substituting a for 

(a - 12)(a - 5) = 0

a = 12        OR         a = 5



a = 12

 ------- Setting  equal to 12



(v - 6)(v + 2) = 0

Therefore, 



a = 5

 ------- Setting  equal to 5



(v - 5)(v + 1) = 0

Therefore,  

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