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

Solution:

Given, the function is f(x) = 2x² + 16x - 9

We have to find the zeros of the quadratic function.

f(x) = 2x² + 16x - 9 = 0

So, 2x² + 16x - 9 = 0

Using quadratic formula,

\(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Here, a = 2, b = 16, c = -9

So, x = \(\frac{-16\pm \sqrt{(16)^{2}-4(2)(-9)}}{2(2)}\)

x = \(\frac{-16\pm \sqrt{256+72}}{4}\\=\frac{-16\pm \sqrt{328}}{4}\)

x = \(\frac{-16\pm 18.11}{4}\)

Now, \(x=\frac{-16+18.11}{4}=\frac{2.11}{4}=0.523\)

\(x=\frac{-16-18.11}{4}=\frac{-34.11}{4}=-8.528\)

Therefore, the zeros of the function are 0.523 and -8.528

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

Summary:

The zeros of the quadratic function f(x) = 2x² + 16x - 9 are 0.523 and -8.528