Question

What are the zeros of the quadratic function f(x) = 2x² + 16x - 9?

Answer 1

The zeros of the quadratic function f(x) = 2x² + 16x - 9 are given as follows:

$x = -4 + \sqrt{\frac{41}{2}}, x = -4 - \sqrt{\frac{41}{2}}$

What is a quadratic function?

A quadratic function is given according to the following rule:

$y = ax^2 + bx + c$

The solutions are:

$x_1 = \frac{-b + \sqrt{\Delta}}{2a}$

$x_2 = \frac{-b - \sqrt{\Delta}}{2a}$

In which:

$\Delta = b^2 - 4ac$

In this problem, the equation is given by:

f(x) = 2x² + 16x - 9.

The coefficients are a = 2, b = 16, c = -9, hence:

$\Delta = 16^2 - 4(2)(-9) = 328$

Then:

$x_1 = \frac{-16 + \sqrt{328}}{2(2)} = -4 + \sqrt{\frac{41}{2}}$

$x_2 = \frac{-16 - \sqrt{328}}{2(2)} = -4 - \sqrt{\frac{41}{2}}$

More can be learned about quadratic functions at brainly.com/question/24737967

#SPJ1