SOLUTION: Hi there, I don't get this problem with quadratic formulas! Could someone please help me work out this problem before class? 2t^3 -6t + 1 = 0 Thank you AC

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Question 17657: Hi there,
I don't get this problem with quadratic formulas! Could someone please help me work out this problem before class?
2t^3 -6t + 1 = 0
Thank you
AC
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
2t^3 -6t + 1 = 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-6x%2B1+=+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=%28-6%29%5E2-4%2A2%2A1=28.

Discriminant d=28 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+28+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+28+%29%29%2F2%5C2+=+2.8228756555323
x%5B2%5D+=+%28-%28-6%29-sqrt%28+28+%29%29%2F2%5C2+=+0.177124344467705

Quadratic expression 2x%5E2%2B-6x%2B1 can be factored:
2x%5E2%2B-6x%2B1+=+%28x-2.8228756555323%29%2A%28x-0.177124344467705%29
Again, the answer is: 2.8228756555323, 0.177124344467705. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-6%2Ax%2B1+%29



But if you're question is:
2t^3 -6t + 1 = 0
Then,
t(2t^2-6)=-1
then t=-1 or (2t^2-6)=-1


(2t^2-6)=-1
2t^2=-1+6
t^2=5/2
+t=sqrt%285%2F2%29+


Then +t=-1 or +t=sqrt%285%2F2%29+


Hope this helps,
Prabhat