Question

what are the zeros of the function $f(x)=x^2 + 5x + 5$ written in simplest radical form?
quadratic formula: $x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$
options:
$x = \frac{5\pm10\sqrt{5}}{2}$
$x = \frac{5\pm\sqrt{5}}{2}$
$x = \frac{-5\pm\sqrt{5}}{2}$
$x = \frac{-5\pm10\sqrt{5}}{2}$

Explanation:

Step1: Identify a, b, c values

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

Step2: Compute discriminant

$\Delta = b^2-4ac = 5^2 - 4(1)(5) = 25 - 20 = 5$

Step3: Apply quadratic formula

$x = \frac{-b\pm\sqrt{\Delta}}{2a} = \frac{-5\pm\sqrt{5}}{2}$

Answer:

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