Question

What are the zeros of the function f(x) = x2 + 8x + 4, expressed in simplest radical form?

x = –4 ± 2StartRoot 3 EndRoot
x = –4 ± StartRoot 48 EndRoot
x = StartFraction negative 8 plus or minus StartRoot 3 EndRoot Over 2 End Fraction
x = StartFraction negative 4 plus or minus 4 StartRoot 3 EndRoot Over 2 End Fraction

Answer 1

To find the zeros of the function \( f(x) = x^2 + 8x + 4 \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = 8 \), and \( c = 4 \).

First, we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 8^2 - 4 \cdot 1 \cdot 4 = 64 - 16 = 48
\]

Now, substituting into the quadratic formula:

\[
x = \frac{-8 \pm \sqrt{48}}{2 \cdot 1}
\]

Next, simplify \( \sqrt{48} \):

\[
\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}
\]

Now substitute this back into the equation:

\[
x = \frac{-8 \pm 4\sqrt{3}}{2}
\]

Now simplify further:

\[
x = \frac{-8}{2} \pm \frac{4\sqrt{3}}{2}
\]
\[
x = -4 \pm 2\sqrt{3}
\]

Thus, the zeros of the function \( f(x) = x^2 + 8x + 4 \) expressed in simplest radical form are:

\[
x = -4 \pm 2\sqrt{3}
\]

Therefore, the corresponding answer from the options is:

\[
x = -4 \pm 2\sqrt{3}
\]