SOLUTION: The value of a stock, A(x), over a year long period decreased and then increased according to the quadratic function 𝐴(𝑥) = 0.75𝑥 2 − 6𝑥 + 20, where x represents t

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: The value of a stock, A(x), over a year long period decreased and then increased according to the quadratic function 𝐴(𝑥) = 0.75𝑥 2 − 6𝑥 + 20, where x represents t      Log On


   



Question 1162485: The value of a stock, A(x), over a year long period decreased and then increased according to the
quadratic function 𝐴(𝑥) = 0.75𝑥
2 − 6𝑥 + 20, where x represents the number of months passed since
you invested. The value of another stock, B(x), increased linearly according to the equation
𝐵(𝑥) = 2.75𝑥 + 1.50 over the same year. After how long are both stocks worth the same amount?

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Set the two equal
0.75x^2-6x+20=2.75x+1.50
0.75x^2-8.75x+18.5=0
x=(1/1.5)(8.75 +/- sqrt (76.56-55.5)); sqrt term is 4.59
the larger root is later in time and is (2/3)(20.22)=8.89 months
(the earlier intersection was at 2.77 months)
can also multiply through by 4 to get 3x^2-35x+74
graph%28300%2C300%2C-2%2C10%2C-10%2C30%2C0.75x%5E2-6x%2B20%2C2.75x%2B1.50%29