Problem 48 - Precalculus Student Solutions Manual 5th

Question

The average cost, in dollars, for tuition and fees for in-state students at four-year public colleges over the period \(2000-2010\) can be modeled by the equation $$y=3.026 x^{2}+377.7 x+3449$$ where \(x=0\) corresponds to 2000 \(x=1\) corresponds to \(2001,\) and so on. Based on this model, for what year after 2000 was the average cost 7605 dollar (Source: The College Board, Anmual Survey of Colleges.) (IMAGE CAN NOT COPY)

Step-by-Step Solution

Answer

The average cost of 7605 dollars was in the year 2010.

Step 1: Identify the given equation

The given quadratic equation that models the average cost for tuition and fees is \[ y = 3.026x^2 + 377.7x + 3449 \] Our goal is to find the value of x for which y equals 7605.

Step 2: Set up the equation

Substitute y with 7605 in the equation to set up the problem: \[ 7605 = 3.026x^2 + 377.7x + 3449 \]

Step 3: Simplify the equation

Subtract 3449 from both sides to simplify the equation: \[ 7605 - 3449 = 3.026x^2 + 377.7x \] This simplifies to: \[ 4156 = 3.026x^2 + 377.7x \]

Step 4: Form a standard quadratic equation

Rearrange the equation to the standard quadratic form: \[ 3.026x^2 + 377.7x - 4156 = 0 \]

Step 5: Solve the quadratic equation

Use the quadratic formula to solve for x, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, a = 3.026, b = 377.7, and c = -4156. Plugging in these values, we get: \[ x = \frac{-377.7 \pm \sqrt{(377.7)^2 - 4(3.026)(-4156)}}{2(3.026)} \]

Step 6: Calculate the discriminant

First, calculate the discriminant ( b^2 - 4ac): \[ (377.7)^2 - 4(3.026)(-4156) = 142655.29 + 50249.456 = 192904.746 \]

Step 7: Find the roots

Now compute the roots using the quadratic formula: \[ x = \frac{-377.7 \pm \sqrt{192904.746}}{2(3.026)} \] This gives us two potential solutions: \[ x_1 = \frac{-377.7 + 439.189}{6.052} = 10.14 \] \[ x_2 = \frac{-377.7 - 439.189}{6.052} = -134.64 \] Since a negative value of x doesn't make sense in this context, we discard it. So we have \( x = 10.14 \).

Step 8: Determine the year

Since x = 0 corresponds to the year 2000, x = 10.14 corresponds to 2000 + 10.14 ≈ 2010 Rounding to the nearest year, it was 2010.

Key Concepts

Tuition Cost Modeling

In this problem, we use a quadratic equation to model the average cost of tuition and fees for in-state students at four-year public colleges. The equation given is: \[ y = 3.026x^2 + 377.7x + 3449 \] Here, 'y' represents the cost, and 'x' corresponds to the years since 2000. This kind of modeling helps in predicting future costs and understanding trends. For instance, setting x = 2 would give an estimate for the year 2002. These models are crucial for planning financial decisions related to education.

Solving Quadratic Equations

To find the year when tuition cost reached a certain value, we need to solve the quadratic equation. Here, we want to determine for which year the average cost was $7605. Thus, we set up the equation as: \[ 7605 = 3.026x^2 + 377.7x + 3449 \] We simplify this to: \[ 3.026x^2 + 377.7x - 4156 = 0 \] Solving quadratic equations involves finding the values of 'x' that satisfy this equation. The options include factoring (when possible), completing the square, or using the quadratic formula.

Quadratic Formula Application

The quadratic formula provides a straightforward method to solve any quadratic equation of the form \(ax^2+bx+c=0\). The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the equation \(3.026x^2 + 377.7x - 4156 = 0\), the coefficients are: * 'a' = 3.026 * 'b' = 377.7 * 'c' = -4156 Plugging these values into the formula, we get: \[ x = \frac{-377.7 \pm \sqrt{(377.7)^2 - 4(3.026)(-4156)}}{2(3.026)} \] This step helps us find the possible values for 'x', which corresponds to the years since 2000.

Discriminant Calculation

An important part of the quadratic formula is the discriminant, given by \(b^2 - 4ac\). The discriminant indicates the nature of the roots: * If positive, there are two distinct real roots. * If zero, there is exactly one real root. * If negative, there are no real roots. In our problem, the discriminant calculation is: \[ (377.7)^2 - 4(3.026)(-4156) = 192904.746 \] Since 192904.746 is positive, we have two potential real roots. Solving further, we obtain: \[ x = \frac{-377.7 \pm 439.189}{6.052} \] This gives us two values, \(x_1 = 10.14\) and \(x_2 = -134.64\). Given that 'x' represents years since 2000, a negative value does not make sense, so we discard it. Hence, \(x = 10.14\) corresponds to the year 2010 when rounded to the nearest whole number.