Answer:
(-5+√5)/2 and (-5-√5)/2
Explanation:
For a general quadratic equation of the form ax²+bx+c = 0
The value of x can be gotten using the general formula expressed as:
x = -b±√b²-4ac/2a
Now given the function
f(x) = x² + 5x + 5
We can get the zeros of the function in its radical form by simply applying the formula above.
From the equation given;
a = 1, b = 5 and c = 5
Substituting this values in the formula we have:
x = -5±√5²-4(1)(5)/2(1)
x = -5±√25-20/2
x = -5±√5/2
x = (-5+√5)/2 and (-5-√5)/2
The zeros of the function in its simplest radical form are
(-5+√5)/2 and (-5-√5)/2