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german
2 years ago
9

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

Mathematics
1 answer:
jeka57 [31]2 years ago
5 0

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}}

<h3>What is a quadratic function?</h3>

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

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