Problem 87 - Contemporary Precalculus

Question

According to data from the U.S. Census Bureau, the population \(P\) of Cleveland, Ohio (in thousands) in year \(x\) can be approximated by \(P=.08 x^{2}-13.08 x+927,\) where \(x=0\) corresponds to \(1950 .\) In what year in the past was the population about \(804,200 ?\)

Step-by-Step Solution

Answer

Answer: The population of Cleveland was approximately 804,200 in the year 1991.

Step 1: Convert the population to thousands

Since our function has the population in thousands, we need to convert the given population of 804,200 to thousands by dividing by 1,000: \(P = \frac{804,200}{1,000} = 804.2\)

Step 2: Plug the value into the equation and solve for x

Substitute the converted population value into the given equation and solve for \(x\): \(804.2 = 0.08x^2 - 13.08x + 927\) This is a quadratic equation, which we will solve to find the value of x. Rearrange the equation by getting all terms on one side of the equation: \(0 = 0.08x^2 - 13.08x + 927 - 804.2\) Simplify the equation: \(0 = 0.08x^2 - 13.08x + 122.8\) Now we can use the quadratic formula to solve for \(x\): \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our equation, \(a=0.08\), \(b=-13.08\), and \(c=122.8\). Plugging the values into the quadratic formula: \(x = \frac{-(-13.08) \pm \sqrt{(-13.08)^2 - 4(0.08)(122.8)}}{2(0.08)}\)

Step 3: Calculate the value of x and find the corresponding year

First, calculate what's inside the square root: \((-13.08)^2 - 4(0.08)(122.8) = 42.6832\) Now calculate the two possible values for \(x\) using the quadratic formula: \(x_1 = \frac{13.08 + \sqrt{42.6832}}{0.16} = 41.427\) \(x_2 = \frac{13.08 - \sqrt{42.6832}}{0.16} = 18.677\) We have two possible values for \(x\). Since the problem asks for the year when this population occurred, we want to use the larger value of \(x\) which corresponds to the population growth. Therefore, we will use \(x_1 = 41.427\). To find the actual year, we add this value to 1950, the base year in our equation: Year = 1950 + 41.427 ≈ 1950 + 41 = 1991 So, the population of Cleveland was approximately 804,200 in the year 1991.

Key Concepts

Solving Quadratic Equations

Understanding how to solve quadratic equations is essential in algebra. These equations are in the standard form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. To solve for \(x\), which represents an unknown variable, one can use various methods such as factoring, completing the square, or the quadratic formula.

Factoring involves breaking down the quadratic equation into a product of simpler binomials. However, this method only works well when the equation easily factors; not all quadratics are factorable. Completing the square is another technique that transforms the quadratic equation into a perfect square trinomial, allowing us to solve for \(x\) by taking the square root of both sides. However, the most universally applicable method is the quadratic formula, a formula that provides the solution to any quadratic equation, regardless of whether it is factorable.

Quadratic Formula

The quadratic formula is a powerful tool that always finds the solutions to any quadratic equation. It is expressed as \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\), where \(a\), \(b\), and \(c\) are the numerical coefficients of the quadratic equation \(ax^2 + bx + c = 0\).

Breaking Down the Formula

The 'b' in the formula relates to the linear coefficient, 'a' to the squared term coefficient, and 'c' to the constant term. The term under the square root, \(b^2 - 4ac\), is known as the discriminant. It determines the nature of the roots. If the discriminant is positive, there are two real and distinct solutions. If it is zero, there is exactly one real solution (also called a repeated or double root). A negative discriminant indicates that there are two complex roots.

When using this formula, one must apply careful attention to the signs and compute both the '+' and '−' versions to obtain the potential two solutions for \(x\).

Population Growth Models

Population growth models use equations to describe how a population changes over time. These models can be linear or exponential, but when the growth is more complex, quadratic or higher-order polynomial equations might be used. In the given exercise, the population of Cleveland, Ohio, is modeled with a quadratic equation.

This model reflects that the population growth does not increase at a constant rate; instead, it may speed up or slow down due to various factors like birth rates, death rates, immigration, and emigration. By applying the quadratic formula to the polynomial equation, one can predict the population for a given year or determine the year when the population reached a specific amount.

These models are not only important for historical data analysis but also for future planning in urban development, resource management, and infrastructure projects.