Answer:
The golf ball was in the air for 4 seconds.
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
In this question:
We have to find the amount of time it takes for the ball to hit the ground. We have that:
[tex]d(t) = -2t^2 + 7t + 4[/tex]
Which is a quadratic equation with [tex]a = -2, b = 7, c = 4[/tex].
How long is the golf ball in the air?
We have to find t for which [tex]d(t) = 0[/tex]
So
[tex]-2t^2 + 7t + 4 = 0[/tex]
[tex]\Delta = b^{2} - 4ac = (7)^2 - 4(-2)(4) = 81[/tex]
[tex]t_{1} = \frac{-7 + \sqrt{81}}{2*(-2)} = -0.5[/tex]
[tex]t_{2} = \frac{-7 - \sqrt{81}}{2*(-2)} = 4[/tex]
Time is a positive measure, so t = 4.
The golf ball was in the air for 4 seconds.
