To find the zeros of the quadratic function f(x) = 16x² + 32x + 9, we need to determine the values of x for which f(x) = 0.
This involves solving the equation:
16x² + 32x + 9 = 0
We can use the quadratic formula, which is given by:
x = (-b ± √(b² - 4ac)) / 2a
Here, a = 16, b = 32, and c = 9. Let’s calculate the discriminant (b² – 4ac) first:
Discriminant = 32² - 4(16)(9) = 1024 - 576 = 448
Since the discriminant is positive, we will have two distinct real zeros. Now we can plug the values into the quadratic formula:
x = (-32 ± √448) / (2 * 16)
Calculating that gives:
x = (-32 ± √448) / 32
To simplify √448, we get:
√448 = √(16 * 28) = 4√28
Substituting this back in gives:
x = (-32 ± 4√28) / 32
This can be further simplified to:
x = -1 ± (√28) / 8
Thus, the zeros of the quadratic function are:
x = -1 + (√28) / 8
x = -1 - (√28) / 8
In conclusion, the zeros of the quadratic function f(x) = 16x² + 32x + 9 are the values we calculated above.