SOLUTION: how do i find all real or imaginary solutions? 3v^2+4v-1=0

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Question 182866: how do i find all real or imaginary solutions?
3v^2+4v-1=0
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
how do i find all real or imaginary solutions?
3v^2+4v-1=0
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Same as always, using the quadratic equation.
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B4x%2B-1+=+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=%284%29%5E2-4%2A3%2A-1=28.

Discriminant d=28 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+28+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+28+%29%29%2F2%5C3+=+0.21525043702153
x%5B2%5D+=+%28-%284%29-sqrt%28+28+%29%29%2F2%5C3+=+-1.54858377035486

Quadratic expression 3x%5E2%2B4x%2B-1 can be factored:
3x%5E2%2B4x%2B-1+=+%28x-0.21525043702153%29%2A%28x--1.54858377035486%29
Again, the answer is: 0.21525043702153, -1.54858377035486. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B4%2Ax%2B-1+%29

The onsite solver's opinion of factoring is a bit loose, and if the coefficient of the x^2 term is not 1, you have to make some adjustments.
Also, it defaults to x for the variable.