SOLUTION: find the roots of 2x^2-5x+1=0 ???? thanx

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Question 26199: find the roots of 2x^2-5x+1=0 ????
thanx
Answer by Alwayscheerful(414) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-5x%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=%28-5%29%5E2-4%2A2%2A1=17.

Discriminant d=17 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+17+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+17+%29%29%2F2%5C2+=+2.28077640640442
x%5B2%5D+=+%28-%28-5%29-sqrt%28+17+%29%29%2F2%5C2+=+0.219223593595585

Quadratic expression 2x%5E2%2B-5x%2B1 can be factored:
2x%5E2%2B-5x%2B1+=+2%28x-2.28077640640442%29%2A%28x-0.219223593595585%29
Again, the answer is: 2.28077640640442, 0.219223593595585. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-5%2Ax%2B1+%29

Hope this helps!