Question

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

A) x = -4- and x = -4+
B) x = 7 and x = 2
C) x = -4- and x = 4-
D) x = -4- and x = -4+, 25
E) x = -4- and x = -4+, 41

Answer 1

The zeros of the quadratic function f(x) = 2x^2 + 16x - 9 are approximately x ≈ 0.54 and x ≈ -8.54, found by applying the quadratic formula.

Step-by-step explanation:

To find the zeros of the quadratic function f(x) = 2x^2 + 16x - 9, we'll need to set the function equal to zero and solve for 'x' using the quadratic formula. The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are coefficients from the quadratic equation ax^2 + bx + c = 0.

For the equation 2x^2 + 16x - 9 = 0, a = 2, b = 16, and c = -9. Plugging these values into the quadratic formula, we get:

x = (-16 ± √((16)^2 - 4(2)(-9))) / (2(2))

x = (-16 ± √(256 + 72)) / 4

x = (-16 ± √(328)) / 4

x = (-16 ± 18.165) / 4

Which gives us two solutions:

• x = (-16 + 18.165) / 4 ≈ 0.54
• x = (-16 - 18.165) / 4 ≈ -8.54

Therefore, the zeros of the function are approximately x ≈ 0.54 and x ≈ -8.54.