SOLUTION: how do i solve the equation? 2x^2+4x+1=0

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Question 448095: how do i solve the equation?
2x^2+4x+1=0
Found 2 solutions by jim_thompson5910, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

2x%5E2%2B4x%2B1=0 Start with the given equation.


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=2, b=4, and c=1


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%284%29+%2B-+sqrt%28+%284%29%5E2-4%282%29%281%29+%29%29%2F%282%282%29%29 Plug in a=2, b=4, and c=1


x+=+%28-4+%2B-+sqrt%28+16-4%282%29%281%29+%29%29%2F%282%282%29%29 Square 4 to get 16.


x+=+%28-4+%2B-+sqrt%28+16-8+%29%29%2F%282%282%29%29 Multiply 4%282%29%281%29 to get 8


x+=+%28-4+%2B-+sqrt%28+8+%29%29%2F%282%282%29%29 Subtract 8 from 16 to get 8


x+=+%28-4+%2B-+sqrt%28+8+%29%29%2F%284%29 Multiply 2 and 2 to get 4.


x+=+%28-4+%2B-+2%2Asqrt%282%29%29%2F%284%29 Simplify the square root


x+=+%28-4%2B2%2Asqrt%282%29%29%2F%284%29 or x+=+%28-4-2%2Asqrt%282%29%29%2F%284%29 Break up the expression.


x+=+%28-2%2Bsqrt%282%29%29%2F%282%29 or x+=+%28-2-sqrt%282%29%29%2F%282%29 Reduce.


So the answers are x+=+%28-2%2Bsqrt%282%29%29%2F%282%29 or x+=+%28-2-sqrt%282%29%29%2F%282%29

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I would use the quadratic equation.