SOLUTION: Find the x-intercepts. y = x^2 + 3x + 1

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Question 104539: Find the x-intercepts.
y = x^2 + 3x + 1
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
The x intercepts occur when y=0
x^2+3x+1=0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B3x%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=%283%29%5E2-4%2A1%2A1=5.

Discriminant d=5 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+5+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+5+%29%29%2F2%5C1+=+-0.381966011250105
x%5B2%5D+=+%28-%283%29-sqrt%28+5+%29%29%2F2%5C1+=+-2.61803398874989

Quadratic expression 1x%5E2%2B3x%2B1 can be factored:
1x%5E2%2B3x%2B1+=+%28x--0.381966011250105%29%2A%28x--2.61803398874989%29
Again, the answer is: -0.381966011250105, -2.61803398874989. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B3%2Ax%2B1+%29