Answer:
the real zeros of the function f(x) = x^2 + 4x + 1 are approximately -2.73 and -1.27.
Explanation:
To find the real zeros of the function f(x) = x^2 + 4x + 1 using the quadratic formula, we can follow these steps:
Step 1: Identify the values of a, b, and c in the quadratic equation, f(x) = ax^2 + bx + c.
In this case, a = 1, b = 4, and c = 1.
Step 2: Write the quadratic formula, which is:
x = (-b ± √(b^2 - 4ac)) / 2a
Step 3: Substitute the values of a, b, and c into the quadratic formula.
x = (-4 ± √(4^2 - 4(1)(1))) / 2(1)
Step 4: Simplify the equation inside the square root.
x = (-4 ± √(16 - 4)) / 2
x = (-4 ± √12) / 2
Step 5: Simplify the expression under the square root.
x = (-4 ± 2√3) / 2
Step 6: Simplify the expression by dividing both the numerator and denominator by 2.
x = -2 ± √3
Therefore, the real zeros of the function f(x) = x^2 + 4x + 1 are approximately -2.73 and -1.27.