To find the zeros of the quadratic function f(x) = 16x² + 32x + 9, we need to solve the equation
16x² + 32x + 9 = 0
We can use the quadratic formula, which is given by:
x = (-b ± √(b² – 4ac)) / 2a
In our case, the coefficients are:
- a = 16
- b = 32
- c = 9
Now, let’s calculate the discriminant (the part under the square root):
b² – 4ac = 32² – 4(16)(9)
Calculating this, we get:
b² – 4ac = 1024 – 576 = 448
Since the discriminant is positive, we will have two distinct real zeros. Now, we can substitute this back into the quadratic formula:
x = (-32 ± √448) / (2 * 16)
Calculating the square root of 448, we find that:
√448 = 4√28
Next, we simplify it:
√448 = 4√(4 * 7) = 4 * 2√7 = 8√7
Substituting this back, we find:
x = (-32 ± 8√7) / 32
We can now separate this into two solutions:
x₁ = (-32 + 8√7) / 32
x₂ = (-32 – 8√7) / 32
To make these solutions a bit clearer:
x₁ = -1 + (√7) / 4
x₂ = -1 – (√7) / 4
Therefore, the zeros of the quadratic function f(x) = 16x² + 32x + 9 are:
- x ≈ -1 + 0.5 ≈ -0.5
- x ≈ -1 – 0.5 ≈ -1.5
In summary, the two zeros of the quadratic function are approximately -0.5 and -1.5.