Problem 82 - Elementary Algebra

Question

A rectangle is 4 in. longer than it is wide. Its area is 117 in. \({ }^{2}\).

Step-by-Step Solution

Answer

The width is 9 inches and the length is 13 inches.

Step 1: Define the variables

Let the width of the rectangle be represented by \( w \). Since the rectangle is 4 inches longer than it is wide, the length can be expressed as \( w + 4 \).

Step 2: Write the area formula

The area of a rectangle is calculated using the formula: \[ \text{Area} = \text{length} \times \text{width} \]. Substituting the given values: \[ 117 = (w + 4) \times w \].

Step 3: Set up the equation

Expand the equation obtained from the area formula: \[ 117 = w^2 + 4w \].

Step 4: Solve the quadratic equation

Rearrange the equation to standard quadratic form: \[ w^2 + 4w - 117 = 0 \]. Solve this quadratic equation using the quadratic formula: \[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1 \), \( b = 4 \), and \( c = -117 \).

Step 5: Calculate the discriminant

Calculate the discriminant: \[ b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-117) = 16 + 468 = 484 \].

Step 6: Find the width, \( w \)

Substitute the discriminant into the quadratic formula: \[ w = \frac{-4 \pm \sqrt{484}}{2 \cdot 1} = \frac{-4 \pm 22}{2} \]. This gives us two potential solutions: \[ w = \frac{18}{2} = 9 \] and \[ w = \frac{-26}{2} = -13 \]. Since width can't be negative, \( w = 9 \).

Step 7: Find the length

The length is 4 inches longer than the width, so \( l = w + 4 = 9 + 4 = 13 \).

Step 8: Verify the solution

Check the area using the found dimensions: \( 9 \times 13 = 117 \), which matches the given area.

Key Concepts

quadratic equations

Quadratic equations form the core of many geometric and algebraic problems. They are written in the standard form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants. In our problem, the quadratic equation arises when we express the area of the rectangle. By setting up the equation \( 117 = w^2 + 4w \), we convert it into standard quadratic form: \[ w^2 + 4w - 117 = 0 \]. Quadratic equations often have two possible solutions. In geometric problems, like finding the dimensions of a rectangle, only positive values are meaningful.

area of a rectangle

The area of a rectangle is given by the product of its length and width. This fundamental formula is written as: \[ \text{Area} = \text{length} \times \text{width} \]. In our exercise, the rectangle's area is 117 square inches. Given one dimension in relation to another, such as 'the length is 4 inches longer than the width,' we use substitution. Writing this mathematically, if \( w \) is the width, then the length \( l = w + 4 \). Substituting these values into the area formula, we get: \[ 117 = (w + 4) \times w \]. This sets up a quadratic equation to be solved.

quadratic formula

The quadratic formula is a powerful tool to solve quadratic equations. It is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. To apply this formula, you need the constants from your equation. For our equation \( w^2 + 4w - 117 = 0 \), the constants are:

• \( a = 1 \)
• \( b = 4 \)
• \( c = -117 \)

. Using these in the quadratic formula helps find the width \( w \). By substituting, we get: \[ w = \frac{-4 \pm \sqrt{484}}{2} \]. This calculation shows why understanding and correctly applying the quadratic formula is crucial in solving such problems.

geometry in algebra

Geometry often intersects with algebra when solving for dimensions and areas. In this problem, we connected algebraic formulas to the geometric property of a rectangle. Here's how:

• Algebra provided the strategy to set up our equation.
• Geometry gave us the context, connecting length and width to the area.

We defined the unknown dimension (width) as \( w \), and expressed the length as \( w + 4 \). This merged geometry into an algebraic equation: \[ (w + 4) \times w = 117 \] which simplifies to: \[ w^2 + 4w - 117 = 0 \]. Understanding both subjects allows you to apply one within the other for solving complex problems.

discriminant calculation

The discriminant of a quadratic equation provides crucial information about the nature of its solutions. It is the part under the square root in the quadratic formula: \[ b^2 - 4ac \]. For our equation \( w^2 + 4w - 117 = 0 \), the discriminant calculation is: \[ 4^2 - 4 \times 1 \times (-117) = 16 + 468 = 484 \]. A positive discriminant indicates two real and distinct solutions. Here, the roots are:

• \( \frac{18}{2} = 9 \) (valid solution)
• \( \frac{-26}{2} = -13 \) (not valid, as width cannot be negative)

. By verifying the geometry constraints, we select the valid solution for our problem-solving context.