SOLUTION: what to solve x^2+8x-4=0 by using the quadratic formula

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Question 500366: what to solve x^2+8x-4=0 by using the quadratic formula
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
what to solve x^2+8x-4=0 by using the quadratic formula
Solve for x, I suppose.
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B8x%2B-4+=+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=%288%29%5E2-4%2A1%2A-4=80.

Discriminant d=80 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-8%2B-sqrt%28+80+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%288%29%2Bsqrt%28+80+%29%29%2F2%5C1+=+0.47213595499958
x%5B2%5D+=+%28-%288%29-sqrt%28+80+%29%29%2F2%5C1+=+-8.47213595499958

Quadratic expression 1x%5E2%2B8x%2B-4 can be factored:
1x%5E2%2B8x%2B-4+=+%28x-0.47213595499958%29%2A%28x--8.47213595499958%29
Again, the answer is: 0.47213595499958, -8.47213595499958. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B8%2Ax%2B-4+%29