Question

what are the zeros of the function $f(x) = x^2 + 5x + 5$ written in simplest radical form?

quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\bigcirc$ $x = \frac{5 \pm 10\sqrt{5}}{2}$

$\bigcirc$ $x = \frac{-5 \pm 10\sqrt{5}}{2}$

$\bigcirc$ $x = \frac{-5 \pm \sqrt{5}}{2}$

$\bigcirc$ $x = \frac{5 \pm \sqrt{5}}{2}$

Explanation:

Step1: Identify coefficients

For \( f(x) = x^2 + 5x + 5 \), \( a = 1 \), \( b = 5 \), \( c = 5 \).

Step2: Apply quadratic formula

Substitute into \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( x = \frac{-5 \pm \sqrt{5^2 - 4(1)(5)}}{2(1)} \)

Step3: Simplify discriminant

Calculate \( 5^2 - 4(1)(5) = 25 - 20 = 5 \).
So \( x = \frac{-5 \pm \sqrt{5}}{2} \).

Answer:

\( x = \frac{-5 \pm \sqrt{5}}{2} \) (corresponding to the option \( x = \frac{-5 \pm \sqrt{5}}{2} \))