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
Algebra ->
Rational-functions
-> 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
Log On
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."
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,
