SOLUTION: I need help with this problem that is for a quadratic unit but I have no idea how to solve it. ((2t)^(-4))-((5t^(-2))+2=0 Thanks

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I need help with this problem that is for a quadratic unit but I have no idea how to solve it. ((2t)^(-4))-((5t^(-2))+2=0 Thanks      Log On


   

Question 148290: I need help with this problem that is for a quadratic unit but I have no idea how to solve it.
((2t)^(-4))-((5t^(-2))+2=0
Thanks
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
2t%5E%28-4%29-5t%5E%28-2%29%2B2=0
This is a quadratic in t%5E%28-2%29
Sub x for t^-2, and it's
2x%5E2-5x-2=0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-5x%2B2+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-5%29%5E2-4%2A2%2A2=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+9+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+9+%29%29%2F2%5C2+=+2
x%5B2%5D+=+%28-%28-5%29-sqrt%28+9+%29%29%2F2%5C2+=+0.5

Quadratic expression 2x%5E2%2B-5x%2B2 can be factored:
2x%5E2%2B-5x%2B2+=+%28x-2%29%2A%28x-0.5%29
Again, the answer is: 2, 0.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-5%2Ax%2B2+%29

x=1/2, x=2
x is t^(-2), so
1%2Ft%5E2=1%2F2
t%5E2+=+2
t=sqrt%282%29,t=-sqrt%282%29
The 2nd root is the inverse, so the other 2 solutions are the inverse of the 2 shown, or
t=%2Bsqrt%282%29%2F2 and t=-sqrt%282%29%2F2