To find the zeros of the function [tex]\( f(x) = x^2 + 5x + 5 \)[/tex], we will use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 5 \)[/tex]
- [tex]\( c = 5 \)[/tex]
First, we will calculate the discriminant, which is the part under the square root in the quadratic formula:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
[tex]\[ \text{Discriminant} = 5^2 - 4(1)(5) \][/tex]
[tex]\[ \text{Discriminant} = 25 - 20 \][/tex]
[tex]\[ \text{Discriminant} = 5 \][/tex]
Now, we can plug the discriminant back into the quadratic formula to find the solutions:
[tex]\[ x = \frac{-b \pm \sqrt{\text{Discriminant}}}{2a} \][/tex]
[tex]\[ x = \frac{-5 \pm \sqrt{5}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-5 \pm \sqrt{5}}{2} \][/tex]
Thus, the zeros of the function [tex]\( f(x) = x^2 + 5x + 5 \)[/tex] in simplest radical form are:
[tex]\[ x = \frac{-5 + \sqrt{5}}{2} \][/tex]
[tex]\[ x = \frac{-5 - \sqrt{5}}{2} \][/tex]
These two values are the solutions to the quadratic equation [tex]\( x^2 + 5x + 5 = 0 \)[/tex].
