SOLUTION: how do you solve x^2+x-5=0 and come out with the answer

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: how do you solve x^2+x-5=0 and come out with the answer       Log On


   



Question 383666: how do you solve x^2+x-5=0 and come out with the answer
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-5+=+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=%281%29%5E2-4%2A1%2A-5=21.

Discriminant d=21 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+21+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+21+%29%29%2F2%5C1+=+1.79128784747792
x%5B2%5D+=+%28-%281%29-sqrt%28+21+%29%29%2F2%5C1+=+-2.79128784747792

Quadratic expression 1x%5E2%2B1x%2B-5 can be factored:
1x%5E2%2B1x%2B-5+=+%28x-1.79128784747792%29%2A%28x--2.79128784747792%29
Again, the answer is: 1.79128784747792, -2.79128784747792. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-5+%29