SOLUTION: Find all solutions, real and imaginary, of the equation: 4t^4

SOLUTION: Find all solutions, real and imaginary, of the equation: 4t^4 - 13t^2 = 12

Question 772794: Find all solutions, real and imaginary, of the equation: 4t^4 - 13t^2 = 12
Found 2 solutions by htmentor, lwsshak3:
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
4t^4 - 13t^2 - 12 = 0
The expression can be factored as (4t^2+3)(t^2-4) = 0
This gives t^2-4 = 0 and 4t^2+3 = 0
t^2 = 4 -> t = 2 and -2
t^2 = -3/4 -> t = sqrt(-3/4) and -sqrt(-3/4) = sqrt(3/4)i and -sqrt(3/4)i
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find all solutions, real and imaginary, of the equation:
4t^4-13t^2=12
4t^4-13t^2-12=0
let u=t^2
u^2=t^4
..
4u^2-13u-12=0
(u-4)(4u+3)=0
u=4=t^2
t^2=�√4=�2
or
u=-3/4=t^2
t^2=�√-3/4=�i√3/2
solutions: -2, 2, -i√3/2, i√3/2
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