SOLUTION: how do u solve this: x^2-4x+1=0

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Question 31989: how do u solve this: x^2-4x+1=0
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^2-4x+1=0
Use the quadratic formula, as follows:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-4x%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=%28-4%29%5E2-4%2A1%2A1=12.

Discriminant d=12 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+12+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+12+%29%29%2F2%5C1+=+3.73205080756888
x%5B2%5D+=+%28-%28-4%29-sqrt%28+12+%29%29%2F2%5C1+=+0.267949192431123

Quadratic expression 1x%5E2%2B-4x%2B1 can be factored:
1x%5E2%2B-4x%2B1+=+1%28x-3.73205080756888%29%2A%28x-0.267949192431123%29
Again, the answer is: 3.73205080756888, 0.267949192431123. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-4%2Ax%2B1+%29

Cheers,
stan H.