To find the real zeros of the quadratic function $f(x) = 9x^2 + 24x + 16$, we need to solve for $x$ when $f(x) = 0$.

So, we set up the equation: $9x^2 + 24x + 16 = 0$

Step 1: Use the Quadratic Formula

The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 9$, $b = 24$, and $c = 16$.

Step 2: Substitute values into the formula

1. Calculate the discriminant $b^2 - 4ac$: $b^2 - 4ac = 24^2 - 4 \cdot 9 \cdot 16 = 576 - 576 = 0$

Since the discriminant is $0$, there is exactly one real root.

2. Substitute into the formula: $x = \frac{-24 \pm \sqrt{0}}{2 \cdot 9} = \frac{-24}{18} = -\frac{4}{3} = -1.33$

Answer

The only real zero is: $x = -1.33$

Final Answer:

• Enter -1.33 in the answer box.

Would you like additional details or have any questions?

Here are some related questions to explore:

1. What happens when the discriminant is greater than zero?
2. How would you solve a quadratic equation with complex roots?
3. What is the significance of rounding in finding real zeros?
4. How does changing the coefficient $a$ affect the position of the zeros?
5. How can you verify the zero using a graphing calculator?

Tip: Always check the discriminant before solving with the quadratic formula; it helps predict the number of real solutions.