What are the zeros of the quadratic function f(x) = 9x² + 54x + 19?

To find the zeros of the quadratic function f(x) = 9x² + 54x + 19, we can use the quadratic formula:

x = (-b ± √(b² – 4ac)) / (2a)

In this equation, a, b, and c are the coefficients from the standard form of a quadratic equation, which is ax² + bx + c = 0.

For our function, we have:

  • a = 9
  • b = 54
  • c = 19

Now, we can substitute these values into the quadratic formula:

x = ( -54 ± √(54² – 4 * 9 * 19)) / (2 * 9)

Calculating the discriminant:

54² = 2916

4 * 9 * 19 = 684

So, the discriminant b² – 4ac = 2916 – 684 = 2232.

Now substituting back into the formula:

x = (-54 ± √2232) / 18

Next, we need to calculate √2232:

√2232 ≈ 47.23 (rounded to two decimal places)

Now we can find the two possible values for x:

x₁ = (-54 + 47.23) / 18 ≈ -0.37

x₂ = (-54 – 47.23) / 18 ≈ -5.67

Therefore, the zeros of the quadratic function f(x) = 9x² + 54x + 19 are approximately:

  • x ≈ -0.37
  • x ≈ -5.67

These values represent the points where the graph of the quadratic function intersects the x-axis.

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