Answer:
One real zero solution, as [tex]\Delta[/tex] is 0, and he must take 12 trips.
Step-by-step explanation:
His amount of funds is given by:
[tex]f(x) = x^2 + 24x + 144[/tex]
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]
How many trips he must take to spend all of his remaining business travel funds?
x trips, and x is found when f(x) = 0. So
[tex]x^2 - 24x + 144 = 0[/tex]
Quadratic equation with [tex]a = 1, b = -24, c = 144[/tex]
So
[tex]\Delta = (-24)^{2} - 4*1*144 = 0[/tex]
[tex]x_{1} = \frac{24 + \sqrt{0}}{2} = 12[/tex]
[tex]x_{2} = \frac{24 - \sqrt{0}}{2} = 12[/tex]
One real zero solution, as [tex]\Delta[/tex] is 0, and he must take 12 trips.