Problem 81 - Elementary Algebra

Question

For exercises 81-86, use a polynomial equation to find the length and width of the rectangle. A rectangle is \(3 \mathrm{in}\). longer than it is wide. Its area is 108 in. \(^{2}\).

Step-by-Step Solution

Answer

The width is 9 inches and the length is 12 inches.

Step 1 - Define Variables

Let the width of the rectangle be denoted as \(w\) inches. Then, the length of the rectangle will be \(w + 3\) inches, since it is given that the length is 3 inches longer than the width.

Step 2 - Write the Area Formula

The area of a rectangle is given by the formula \[ \text{Area} = \text{length} \times \text{width} \]. Substituting the known values, we get: \[ 108 = w \times (w + 3) \]

Step 3 - Form the Polynomial Equation

Expand and rearrange the equation to form a polynomial: \[ 108 = w \times (w + 3) \] \[ 108 = w^2 + 3w \] \[ w^2 + 3w - 108 = 0 \]

Step 4 - Solve the Polynomial Equation

Solve the quadratic equation \[ w^2 + 3w - 108 = 0 \] using the quadratic formula \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Here, \a = 1\, \b = 3\, and \c = -108\. Substitute these values into the formula: \[ w = \frac{-3 \pm \sqrt{3^2 - 4(1)(-108)}}{2(1)} \] \[ w = \frac{-3 \pm \sqrt{9 + 432}}{2} \] \[ w = \frac{-3 \pm \sqrt{441}}{2} \] \[ w = \frac{-3 \pm 21}{2} \]

Step 5 - Determine the Width

Calculate the potential solutions for the width: \[ w = \frac{-3 + 21}{2} = 9 \] and \[ w = \frac{-3 - 21}{2} = -12 \]. Since width cannot be negative, we select \w = 9\ inches.

Step 6 - Calculate the Length

Use the width to find the length: \[ \text{Length} = w + 3 = 9 + 3 = 12 \text{inches} \]

Key Concepts

Solving Quadratic Equations

In many geometric problems, we end up with quadratic equations, especially those involving areas of rectangles. A quadratic equation is any equation of the form \[ ax^2 + bx + c = 0 \]. Solving these requires special strategies like factoring, completing the square, or using the quadratic formula. In our example, the quadratic equation is \[ w^2 + 3w - 108 = 0 \], where \( a = 1 \), \( b = 3 \), and \( c = -108 \).Breaking down the quadratic formula step-by-step can be very helpful:

• The quadratic formula is \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
• First, calculate the discriminant \( b^2 - 4ac \). In our case, it is \( 3^2 - 4(1)(-108) = 9 + 432 = 441 \).
• Next, take the square root of the discriminant: \( \sqrt{441} = 21 \).
• Finally, apply these values in the formula: \( w = \frac{-3 \pm 21}{2} \), giving us two potential solutions.
• Check both solutions to find which one is valid in the context of the problem. Since width cannot be negative, we select \( w = 9 \) inches.

This systematic approach simplifies the problem and arrives at accurate results.

Area of a Rectangle

The area of a rectangle is a fundamental concept in geometry. It represents the amount of space inside the rectangle's boundaries. The formula for the area is straightforward: \[ \text{Area} = \text{length} \times \text{width} \].In our problem, the area is given as 108 square inches. By defining the width as \( w \) inches and the length as \( w + 3 \) inches, we can set up an equation with these variables: \[ 108 = w \times (w + 3) \]. This shows us that the area calculation is a direct multiplication of the two dimensions of the rectangle.Understanding this concept helps in forming the polynomial equation needed to solve for the rectangle's dimensions. Once you interpret the area formula into a polynomial equation, the challenge then becomes a matter of algebra rather than geometry. This approach is useful for various problems, not just rectangles, making it a valuable mathematical technique.

Variable Substitution

Variable substitution is a powerful technique in algebra and geometry. It involves replacing complex expressions with simpler variables to make manipulation easier. In our problem, we used variable substitution to simplify the complexity of the geometric shape. Here’s how it was done:

• First, identify the unknown quantities. In this case, the lengths of the rectangle’s sides.
• Then, choose a variable to represent one of these unknowns. We chose \( w \) for the width.
• Express other unknowns in terms of this variable. The length, for example, is \( w + 3 \) since it is 3 inches longer than the width.
• Substitute these variables back into the area formula to form an equation. \[ 108 = w \times (w + 3) \], which simplifies to the quadratic equation \( w^2 + 3w - 108 = 0 \).

By substituting variables in this manner, the problem shifts from a geometric context to a purely algebraic one, making it easier to apply algebraic techniques like solving quadratic equations. This method is widely applicable in various subjects and problem types.