Problem 52 - College Algebra

Question

Abercrombie and Fitch stock had a price given as \(P=0.2 t^{2}-5.6 t+50.2,\) where \(t\) is the time in months from 1999 to \(2001 .(t=1 \text { is January } 1999) .\) Find the two months in which the price of the stock was \(\$ 30 .\)

Step-by-Step Solution

Answer

The stock price was $30 in April 1999 and January 2001.

Step 1: Set the Equation Equal to 30

The goal is to find the values of \( t \) (months) when the stock price \( P \) is \( 30 \). We start by setting the equation \( P=0.2t^2-5.6t+50.2 \) equal to 30: \[ 0.2t^2 - 5.6t + 50.2 = 30 \]

Step 2: Simplify the Equation

Subtract 30 from both sides to simplify the equation: \[ 0.2t^2 - 5.6t + 20.2 = 0 \]

Step 3: Solve the Quadratic Equation

We have a quadratic equation in the form \( at^2 + bt + c = 0 \). Here, \( a = 0.2 \), \( b = -5.6 \), and \( c = 20.2 \). Use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Calculate the discriminant: \( b^2 - 4ac = (-5.6)^2 - 4 \times 0.2 \times 20.2 \). This results in a discriminant of \( 31.36 - 16.16 = 15.2 \).

Step 4: Calculate the Roots

Using the quadratic formula: \[ t = \frac{-(-5.6) \pm \sqrt{15.2}}{2 \times 0.2} = \frac{5.6 \pm \sqrt{15.2}}{0.4} \]. Let's compute the two roots: \( t_1 = \frac{5.6 + \sqrt{15.2}}{0.4} \) and \( t_2 = \frac{5.6 - \sqrt{15.2}}{0.4} \). Calculating these, we get two solutions for \( t \).

Step 5: Verify the Solutions

Compute \( \sqrt{15.2} \), which approximately equals \( 3.8987 \). Substitute into the formulas to find: \[ t_1 = \frac{5.6 + 3.8987}{0.4} \approx \frac{9.4987}{0.4} \approx 23.75 \] and \[ t_2 = \frac{5.6 - 3.8987}{0.4} \approx \frac{1.7013}{0.4} \approx 4.253 \]. We round these to 24 and 4 (since \( t \) represents complete months).

Step 6: Confirm the Months

The solutions correspond to \( t = 24 \) (January, 2001) and \( t = 4 \) (April, 1999). Check this by plugging these values back into the original equation to confirm \( P=30 \) for both months.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \( ax^2 + bx + c = 0 \). This formula helps you find the values of \( x \) (or any variable used) where the quadratic equation equals zero. The formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this equation, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation.

• \( a \): the coefficient of \( x^2 \)
• \( b \): the coefficient of \( x \)
• \( c \): the constant term

The \( \pm \) sign indicates that there are typically two solutions, corresponding to the addition and subtraction of the square root term. These solutions are also known as the "roots" of the equation.

Stock Price

A stock price is the current price at which a particular stock can be bought or sold in the market. In the given problem, the stock price \( P \) of Abercrombie and Fitch is represented by the quadratic equation \( P = 0.2t^2 - 5.6t + 50.2 \), where \( t \) is the time in months. This equation shows how the stock price changes over time.

• Stock prices can fluctuate due to economic factors, company performance, market trends, and investor sentiment.
• Predicting stock price movements can be complex due to several influencing factors.

In the exercise, the task is to find the months when the stock price was exactly $30 by solving the quadratic equation using \( t \). This demonstrates one way mathematical models can help analyze and predict stock trends.

Time in Months

Time in months is a crucial variable in this stock price problem. Here, it is represented by \( t \) and acts as a sequence of intervals from January 1999 onward.

• \( t = 1 \) corresponds to January 1999.
• As \( t \) increases by one unit, it progresses to the next month.
• The problem requires calculating specific months (\( t \)) when certain conditions are met, like the stock price being \(30.

By substituting values of \( t \) in the stock price equation, one can determine how the price behaves over different months. Solving for \( t \) when the stock price equals \)30 identifies the specific months of interest, enhancing the understanding of this timeline and pattern recognition.

Discriminant

The discriminant is a part of the quadratic formula that helps determine the nature of the roots of a quadratic equation. It is found under the square root in the quadratic formula: \( b^2 - 4ac \).

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it is zero, the equation has exactly one real root (repeated root).
• If it is negative, the equation has no real roots, but rather two complex roots.

In the exercise, the discriminant was calculated as 15.2, which is positive. This tells us there are two real and distinct solutions for \( t \). Understanding the discriminant aids in predicting the behavior of the roots without solving the entire equation.