Problem 84 - Elementary Algebra

Question

The width of a rectangle is \(7 \mathrm{ft}\) less than its length. Its area is \(120 \mathrm{ft}^{2}\).

Step-by-Step Solution

Answer

The length is 15 ft and the width is 8 ft.

Step 1 - Define Variables

Let the length of the rectangle be represented by the variable \(L\). Since the width is \(7 \mathrm{ft}\) less than the length, the width can be represented as \(L - 7\).

Step 2 - Write the Area Formula

The area of a rectangle is given by the formula \(A = L \times W\), where \(A\) is the area, \(L\) is the length, and \(W\) is the width. Substitute the values and the expression for the width into the equation: \[120 = L \times (L - 7)\]

Step 3 - Expand the Equation

Expand the right side of the equation: \[120 = L^2 - 7L\]

Step 4 - Form a Quadratic Equation

Rearrange the equation to form a standard quadratic equation: \[L^2 - 7L - 120 = 0\]

Step 5 - Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula \(L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -7\), and \(c = -120\). Compute the discriminant \(b^2 - 4ac\): \[-(7)^2 - 4(1)(-120) = 49 + 480 = 529\]. Now find the values of \(L\): \[L = \frac{7 \pm \sqrt{529}}{2} = \frac{7 \pm 23}{2}\].

Step 6 - Determine the Length

Calculate the two potential solutions: \[L = \frac{7 + 23}{2} = 15 \ \text{or} \ L = \frac{7 - 23}{2} = -8\]. Disregard the negative solution since a length cannot be negative. Thus, the length is \(15 \mathrm{ft}\).

Step 7 - Find the Width

Use the expression for the width: \[W = L - 7 \rightarrow W = 15 - 7 = 8\ \mathrm{ft}\]

Key Concepts

defining variables

Understanding how to define variables is crucial in solving algebraic problems. Variables are symbols that represent unknown values. In this problem, we needed to find the length and width of a rectangle. Therefore, we used a variable to represent one of these measurements.
Let's take the length of the rectangle as our variable, represented by the letter \(L\). Since the width is 7 feet less than the length, we express the width as \(L - 7\). Using variables simplifies the process of forming equations that describe the relationships within the problem.
This step sets the foundation for forming and solving the equation in subsequent steps.

area of a rectangle

The area of a rectangle is the space contained within its sides. It's calculated using the formula: \(A = L \times W\), where \(A\) is the area, \(L\) is the length, and \(W\) is the width.
In our problem, we are given the area, 120 square feet, and expressions for the length and width. When we substitute these values and expressions into our area formula, we get:

\t
• \(120 = L \times (L - 7)\): We replace \(A\) with 120 and \(W\) with \(L - 7\).

Expanding this equation helps us to express it as a quadratic equation, which is easier to solve.

quadratic formula

The quadratic formula is essential for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is: \(L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
In our problem, we transformed our area equation into a quadratic equation:

\t
• \(L^2 - 7L - 120 = 0\): Here, \(a = 1\), \(b = -7\), and \(c = -120\).

We then use the quadratic formula:

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• Calculate the discriminant: \t\t
  \t\t\t
  • \(b^2 - 4ac = (-7)^2 - 4(1)(-120) = 49 + 480 = 529\)
  \t\t
  \t
\t
• Then, plug in the values to find \(L\): \t\t
  \t\t\t
  • \(L = \frac{7 \pm \sqrt{529}}{2} = \frac{7 \pm 23}{2}\)
  \t\t\t
  • This gives us two potential solutions:\(L = 15\) or \(L = -8\). Since lengths can’t be negative, we select \(L = 15\).
  \t\t
  \t

Finally, we determine the width using \(W = L - 7\), giving us 8 feet for the width. This method ensures we account for all possible values and choose the ones that make sense in the real world.