SOLUTION: Anybody know this one x^2+4x+1>0 thanks for your time

Question 62029: Anybody know this one
x^2+4x+1>0
thanks for your time
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
x^2 + 4x + 1 > 0
Vertex ~> (-b/2a,f(x))
Vertex ~> (-2,-3)
The vertex is located under the x-axis and in the third quadrant....where will the parabola hit the x-axis (x value is zero) also known as the x-intercept?
x^2 + 4x + 1 = 0
x^2 + 4x = -1
(x + 2)^2 = 3
x + 2 = +- sqrt(3)
x = -2 +- sqrt(3) You can also use the quadratic formula....
We have:
<------(-2 - sqrt(3))----vertex----(-2 + sqrt(3))------->
We have:
<======(-2 - sqrt(3))----vertex----(-2 + sqrt(3))=======>
(-,-2 - sqrt(3))U(-2 + sqrt(3),+)

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