To find the zeros of the function f(x) = x² + 5x + 5, we can use the quadratic formula, which is given by:
x = (-b ± sqrt(b² – 4ac)) / 2a
In this case, the coefficients are:
- a = 1 (the coefficient of x²)
- b = 5 (the coefficient of x)
- c = 5 (the constant term)
Now, let’s calculate the discriminant, which is b² – 4ac:
Discriminant = 5² – 4(1)(5) = 25 – 20 = 5
Since the discriminant is positive, we will have two real and distinct solutions. Next, we substitute the values into the quadratic formula:
x = (-5 ± sqrt(5)) / (2 * 1)
This simplifies to:
x = (-5 ± sqrt(5)) / 2
Therefore, the zeros of the function f(x) = x² + 5x + 5 are:
- x = (-5 + sqrt(5)) / 2
- x = (-5 – sqrt(5)) / 2
To express these in simplest radical form:
1. The first zero is x1 = (-5 + sqrt(5)) / 2
2. The second zero is x2 = (-5 – sqrt(5)) / 2
In conclusion, the zeros of the function f(x) = x² + 5x + 5 written in simplest radical form are x = (-5 ± sqrt(5)) / 2.