Final answer:
The real zeros of the quadratic function f(x) = x^2 + 4x + 1 are approximately -2 + sqrt(3) and -2 - sqrt(3).
Step-by-step explanation:
To find the real zeros of the quadratic function f(x) = x^2 + 4x + 1, we can start by factoring the quadratic equation or using the quadratic formula. In this case, we will use the quadratic formula.
The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
Substituting the values of a, b, and c from the given function f(x) = x^2 + 4x + 1, we can calculate the real zeros:
x = (-4 ± sqrt(16 - 4(1)(1))) / (2(1))
Simplifying further:
x = (-4 ± sqrt(12)) / (2)
x = (-4 ± sqrt(4*3)) / (2)
x = (-4 ± 2*sqrt(3)) / (2)
x = -2 ± sqrt(3)
Therefore, the real zeros of the quadratic function f(x) = x^2 + 4x + 1 are approximately:
x ≈ -2 + sqrt(3)
x ≈ -2 - sqrt(3)