Question

The price C, in dollars per share, of a high-tech stock has fluctuated over a twelve-year period according to the equation C= 14 +12x – x2, where x is in years. The price C, in dollars per share, of a second high-tech stock has shown a steady increase during the same time period according to the relationship C = 2x + 30. For what values are the two stock prices the same?

Answer 1

The values for which the two stock prices are the same are approximately x ≈ -1.405 and x ≈ 11.405.

Step-by-step explanation:

To find the values for which the two stock prices are the same, we need to set the equations for the prices equal to each other and solve for x.

Equation for the first stock: C = 14 + 12x - x^2

Equation for the second stock: C = 2x + 30

Setting the two equations equal: 14 + 12x - x^2 = 2x + 30

Rearranging the equation and combining like terms: x^2 - 10x - 16 = 0

Using the quadratic formula to solve for x: x = (-b ± sqrt(b^2 - 4ac))/(2a)

Plugging in the values: x = (-(-10) ± sqrt((-10)^2 - 4(1)(-16)))/(2(1))

Simplifying: x = (10 ± sqrt(100 + 64))/2

Calculating: x = (10 ± sqrt(164))/2

Approximate values: x ≈ (10 ± 12.81)/2

Therefore, the two stock prices are the same for x ≈ -1.405 and x ≈ 11.405.

Answer 2

For this case we have the following equations:
C = 14 + 12x - x2
C = 2x + 30
Equating the equations we have:
14 + 12x - x2 = 2x + 30
Rewriting we have:
-x2 + 10x - 16 = 0
Solving the polynomial we have:
x1 = 2
x2 = 8
Answer:
the two stock prices are the same for:
x1 = 2
x2 = 8