To find the zeros of the function f(x) = x² + 8x + 4, we need to solve the equation f(x) = 0. This means we will set the quadratic equation to zero:
x² + 8x + 4 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b² – 4ac)) / (2a)
In our function, the coefficients are:
- a = 1 (the coefficient of x²)
- b = 8 (the coefficient of x)
- c = 4 (the constant term)
Now we can substitute these values into the quadratic formula:
x = (-(8) ± √((8)² – 4(1)(4))) / (2(1))
This simplifies to:
x = (-8 ± √(64 – 16)) / 2
x = (-8 ± √48) / 2
Next, we simplify the square root:
√48 = √(16 * 3) = 4√3
Now, substituting back into the equation gives us:
x = (-8 ± 4√3) / 2
We can further simplify this by dividing both the terms in the numerator by 2:
x = -4 ± 2√3
Thus, the two zeros of the function f(x) = x² + 8x + 4 in simplest radical form are:
x = -4 + 2√3 and x = -4 – 2√3.