Problem 76 - Introductory Algebra for College Students

Question

The alligator, at one time an endangered species, is the subject of a protection program. The formula $$P=-10 x^{2}+475 x+3500$$ models the alligator population, \(P,\) after \(x\) years of the protection program, where \(0 \leq x \leq 12 .\) Use the formula to solve Exercises \(76-77\) After how long is the population up to \(5990 ?\)

Step-by-Step Solution

Answer

After approximately 3.15 years, the population of the alligators will increase to 5990.

Step 1: Set up the equation

First, substitute \(P\) in the given equation with \(5990\) and then solve for \(x\). This produces the equation: \(-10x^{2} + 475x + 3500 = 5990\)

Step 2: Simplify equation

The equation simplifies to: \(-10x^{2} + 475x - 2490 = 0\). Now that we have a quadratic equation, we can compute the values of \(a, b,\) and \(c\). Here, \(a = -10, b = 475\) and \(c = -2490\).

Step 3: Solving the quadratic equation

Substitute \(a, b,\) and \(c\) into the quadratic formula to solve for \(x\). Plugging these values in gives us \(x = \frac{-475 \pm \sqrt{475^2-4*(-10)*(-2490)}}{2*(-10)}\). This simplifies further to \(x = \frac{-475 \pm \sqrt{169750}}{-20}\). The sqrt of \(169750\) is approximately \(412.00\). Thus, we have two possible solutions for \(x\), \(x = \frac{-475 + 412}{-20}\) and \(x = \frac{-475 - 412}{-20}\).

Step 4: Validating the Answers

The first solution gives \(x = -3.15\) and the second gives \(x = 44.35\). However, as per the given constraints in the problem \(0 \leq x \leq 12\), only \(x = -3.15\) is valid.

Key Concepts

Algebraic Modeling

Algebraic modeling is a method of using mathematical equations to represent real-world situations. In this context, an equation is created to simulate the population growth of alligators over time within a protected environment. Specifically, the alligator protection program's effectiveness is expressed through the quadratic equation \(P = -10x^2 + 475x + 3500\), where \(P\) represents the population after \(x\) years. This model allows us to predict and analyze the outcomes under different scenarios by substituting values and solving for the unknowns.

Understanding the components of this model is crucial. The equation is structured in a way that the term \(x^2\) indicates the rate of change in population growth is not constant, but affected by the time variable \(x\) squared. The coefficient \(–10\) reveals that growth rate decreases over time, a realistic aspect since populations can't grow indefinitely. Moreover, modeling helps identify when certain population milestones are reached, such as the \(P = 5990\) in the exercise, by providing a concrete framework to mathematically solve for these events.

Solving Quadratic Equations

Solving quadratic equations is a fundamental aspect of algebra that involves finding the value(s) of the variable that satisfy the equation \(ax^2 + bx + c = 0\). There are various methods to solve them, such as factoring, completing the square, and the quadratic formula. The exercise employs the quadratic formula, a universal method especially useful when other techniques are inconvenient or inapplicable.

The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation, and \(\pm\) indicates that typically two solutions will emerge—a positive and a negative root. In this exercise, after substituting the coefficients into the formula, simplifying, and calculating the discriminant \(\sqrt{b^2-4ac}\), we find two potential solutions. However, based on the initial problem constraints, we must discard any solution that does not fall within the specified range of \(0 \leq x \leq 12\). The concept of solution validation is an essential part of solving quadratic equations as it ensures that the solutions make sense in the given context.

Mathematical Problem-Solving

Mathematical problem-solving encompasses understanding the problem, devising a plan, carrying out the plan, and reflecting upon the solution. It is a critical skill in mathematics that extends beyond the confines of the subject into practical, real-life applications.

In our problem, understanding the situation involved reading and interpreting the given equation for the alligator population. The plan included setting up and simplifying the equation to solve for the time it takes the population to reach 5990. Upon substituting the known values, the action phase involved applying algebraic techniques and the quadratic formula to find the solution. The final and often overlooked stage is reflecting upon and validating results against given constraints—only one of the calculated solutions was within the allowable range. This problem-solving process not only yields the correct result but also reinforces a systematic approach that students can replicate across diverse mathematical challenges.