Problem 29 - Elementary Algebra

Question

Solve each quadratic equation by completing the square. A study of the air quality by an environmental group suggests that \(t\) years from now the level of carbon monoxide in the air, in parts per million, will be given by the quadratic equation $$ A=0.4 t^{2}+0.1 t+3.1 $$ where \(A\) represents the amount of carbon monoxide in the air. (a) What is the level, in parts per million, of carbon monoxide in the air now? (b) How many years from now will the level of carbon monoxide be at 18.1 parts per million?

Step-by-Step Solution

Answer

Answer: The current level of carbon monoxide in the air is 3.1 parts per million, and it will reach 18.1 parts per million in approximately 5.30 years from now.

Step 1: Find the current level of carbon monoxide (A) when t=0

To find the current level of carbon monoxide, we need to substitute t=0 in the given equation and solve for A. $$ A = 0.4t^2 + 0.1t + 3.1 $$ Substitute t=0: $$ A = 0.4(0)^2 + 0.1(0) + 3.1 $$ Simplify: $$ A = 0 + 0 + 3.1 $$ The current level of carbon monoxide in the air is 3.1 parts per million.

Step 2: Find the time (t) when the level of carbon monoxide is 18.1 parts per million

To find the time when the level of carbon monoxide reaches 18.1 parts per million, we need to substitute A=18.1 in the given equation and solve for t. $$ A = 0.4t^2 + 0.1t + 3.1 $$ Substitute A=18.1: $$ 18.1 = 0.4t^2 + 0.1t + 3.1 $$ We need to solve the quadratic equation for t. First, we can subtract 18.1 from both sides to move all the terms to one side and simplify: $$ 0 = 0.4t^2 + 0.1t - 15 $$ We will solve this quadratic equation using the quadratic formula: $$ t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ In our equation, a=0.4, b=0.1, and c=-15. Substitute the values and solve for t: $$ t=\frac{-0.1\pm\sqrt{(0.1)^2-4(0.4)(-15)}}{2(0.4)} $$ Simplify: $$ t=\frac{-0.1\pm\sqrt{0.01+24}}{0.8} $$ $$ t=\frac{-0.1\pm\sqrt{24.01}}{0.8} $$ Now we have two possible solutions for t: $$ t_1=\frac{-0.1+\sqrt{24.01}}{0.8} $$ $$ t_2=\frac{-0.1-\sqrt{24.01}}{0.8} $$ Calculate the values of \(t_1\) and \(t_2\): $$ t_1 \approx 5.30 $$ $$ t_2 \approx -5.62 $$ Since we are measuring time in years, a negative value for t doesn't make sense in this context. Therefore, the level of carbon monoxide will be at 18.1 parts per million in approximately 5.30 years from now. The current level of carbon monoxide in the air is 3.1 parts per million (a). The level will reach 18.1 parts per million in approximately 5.30 years from now (b).

Key Concepts

Quadratic Equations

Quadratic equations are expressions where the highest power of an unknown variable, usually represented as \(t\) or \(x\), is two. This forms a standard structure represented by the equation: \[ax^2 + bx + c = 0\] where \(a\), \(b\), and \(c\) are coefficients, with \(a eq 0\). This specific exercise involves the quadratic equation \[A = 0.4t^2 + 0.1t + 3.1\] used to model the carbon monoxide level. Quadratic equations like these are common in many real-life applications including physics, finance, and environmental studies. They describe a parabolic relationship, indicating a peak or a trough, depending on the coefficients involved.When solving quadratic equations, you can employ various methods such as:

• Factorization
• Completing the square
• Using the quadratic formula

Understanding how to manipulate and solve these equations is crucial when trying to predict outcomes based on mathematical models.

Carbon Monoxide Level

The exercise's quadratic equation models carbon monoxide level changes over time. Carbon monoxide (CO) is a colorless, odorless gas that can be harmful when inhaled. It's essential to understand the levels of CO in the environment, especially in urban areas where pollution is higher due to vehicle emissions and industrial activities.In the provided exercise, the equation \[A = 0.4t^2 + 0.1t + 3.1\] represents the CO level in parts per million (ppm), where \(A\) denotes the amount of CO, and \(t\) represents the time in years. Understanding the initial value of \(A\) when \(t=0\), which is 3.1 ppm, helps gauge the present carbon monoxide levels. As time progresses, the prediction model helps assess how CO levels might increase reaching certain thresholds, like 18.1 ppm. Monitoring changes over time with such quadratic models enables scientists and policymakers to anticipate possible health risks and implement necessary measures to reduce pollution.

Solving Equations

Solving equations is about finding the value(s) of the unknowns that make the equation true once substituted back into it. In our case, the main goal is to find \(t\), the point in time when specific conditions about the carbon monoxide levels occur, using the quadratic equation.The exercise focuses on determining \(t\) when \(A = 18.1\) ppm:\[18.1 = 0.4t^2 + 0.1t + 3.1\]Start by rearranging:\[0 = 0.4t^2 + 0.1t - 15\]The quadratic formula provides a reliable method when equations are not easily factored:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Substituting the values from the equation (\(a=0.4\), \(b=0.1\), \(c=-15\)) allows solving for two possible \(t\) values. Given the context, we interpret \(t\) positively and discard negative solutions, leading to the conclusion the level reaches 18.1 in about 5.30 years.By applying these steps, one learns how to tackle various quadratic problems across different fields, offering an intuitive understanding of how changes over time can be modeled and predicted. This process amplifies problem-solving skills crucial for tackling real-world data-related challenges.