Problem 83 - Algebra for College Students

Question

Find the exact solution \((s)\) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. Area of a rectangle. The length of a rectangle is \(4 \mathrm{ft}\) longer than the width, and its area is 10 square feet ( \(\mathrm{ft}^{2}\) ). Find the length and width.

Step-by-Step Solution

Answer

Width: \sqrt{14} - 2 \approx 1.7 ft, Length: \sqrt{14} + 2 \approx 5.7 ft

Step 1: Define variables

Let the width of the rectangle be \(w\) feet. Then the length of the rectangle is \(w + 4\) feet.

Step 2: Set up the equation for area

The area of the rectangle is given by the product of its length and width. Thus, \(w \times (w + 4) = 10\).

Step 3: Expand and simplify

Expand and simplify the equation: \(w^2 + 4w - 10 = 0\).

Step 4: Solve the quadratic equation

Use the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 4\), and \(c = -10\). Plugging in the values, \(w = \frac{-4 \pm \sqrt{16 + 40}}{2} = \frac{-4 \pm \sqrt{56}}{2}\).

Step 5: Simplify the solution

Simplify the expression to get: \(w = \frac{-4 \pm 2\sqrt{14}}{2}\), which further simplifies to \(w = -2 \pm \sqrt{14}\). Since width cannot be negative, \(w = \sqrt{14} - 2\).

Step 6: Find the length

The length is given by \(w + 4\). Substituting \(w = \sqrt{14} - 2\), the length is \(w + 4 = \sqrt{14} + 2\).

Step 7: Approximate the solution

Approximate \(\sqrt{14} \approx 3.7\), thus the width is \(3.7 - 2 = 1.7\) feet and the length is \(3.7 + 2 = 5.7\) feet.

Key Concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2+bx+c=0\). This equation surfaces frequently in algebra and beyond. It allows us to find the values of the variable - let's say w in our problem - that make the equation true. Setting it up requires identifying the coefficients (\(a, b, c\)) from the standard form of the equation. In our case, we have \(w^2 + 4w - 10 = 0\), which leads us to identify a = 1, b = 4, and c = -10.
To solve for w, we use the quadratic formula:
\( x = \frac{-b \textpm \textsqrt{b^2-4ac}}{2a} \). Plugging in our values, we get \( w = \frac{-4 ± \textsqrt{56}}{2} \). Simplifying this further results in \( w = -2 ± \textsqrt{14} \).
Since the width cannot be negative, we discard the negative solution. So, the width (\(w\)) becomes \( sqrt{14} - 2 \).
The quadratic formula not only finds real solutions but also complex ones, making it very versatile.

area of a rectangle

The area of a rectangle is a fundamental concept in geometry. It is calculated by multiplying the length by the width:
\( A = l \times w \). In our problem, we know that the area is 10 square feet, and the length is 4 feet longer than the width. Representing the width as w and the length as w + 4, our area equation becomes:
\( w(w + 4) = 10 \). Expanding this, we get \( w^2 + 4w - 10 = 0 \), a setup for a quadratic equation.
We solve this equation to find possible values for w. Once we know the width, adding 4 gives us the length. This interdependence clearly shows the importance of properly setting up and solving equations in geometry.

irrational numbers

Irrational numbers are numbers that cannot be expressed as a simple fraction. They have non-repeating, non-terminating decimal expansions.
Examples include \(\textsqrt{2}, \textpi\), and in our problem, \(\text\textsqrt{14}\).
We encountered \(\textsqrt{14}\) when solving using the quadratic formula.
Since it's irrational, we cannot write it precisely as a simple fraction. We instead use an approximate value.
Here, \(\textsqrt{14} \text\textapprox 3.7\).
This gave us approximate physical dimensions for the rectangle, ensuring practical understanding.
Recognizing and working with irrational numbers is essential in many areas of math and science.