Math Problem Statement

A person has a garden that has a length 4 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 60

ft^2. Solve the quadratic equation to find the length and width.

Enter the exact answers.

Solution

Let's break down the information given:

  1. Let the width of the garden be xx feet.
  2. Since the length is 4 feet longer than the width, the length will be x+4x + 4 feet.
  3. The area of the garden is given as 60 square feet.

Using the formula for the area of a rectangle, which is: Area=length×width\text{Area} = \text{length} \times \text{width}

we can set up the equation: x(x+4)=60x(x + 4) = 60

Step 1: Set up the quadratic equation

Expanding the equation: x2+4x=60x^2 + 4x = 60

Then, rearrange it to bring all terms to one side: x2+4x60=0x^2 + 4x - 60 = 0

Step 2: Solve the quadratic equation

Now we have a quadratic equation: x2+4x60=0x^2 + 4x - 60 = 0

To solve for xx, we can use the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

where a=1a = 1, b=4b = 4, and c=60c = -60.

Substitute values:

x=4±4241(60)21x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-60)}}{2 \cdot 1} x=4±16+2402x = \frac{-4 \pm \sqrt{16 + 240}}{2} x=4±2562x = \frac{-4 \pm \sqrt{256}}{2} x=4±162x = \frac{-4 \pm 16}{2}

Calculate the two possible values of xx:

  1. x=4+162=122=6x = \frac{-4 + 16}{2} = \frac{12}{2} = 6
  2. x=4162=202=10x = \frac{-4 - 16}{2} = \frac{-20}{2} = -10 (We discard this negative value because dimensions cannot be negative.)

So, the width xx is 6 feet.

Step 3: Find the length

Since the length is x+4x + 4: Length=6+4=10 feet\text{Length} = 6 + 4 = 10 \text{ feet}

Final Answer:

The dimensions of the garden are:

  • Width: 6 feet
  • Length: 10 feet

Would you like more details or have any questions?


Here are five related questions to deepen your understanding:

  1. What would the quadratic equation be if the area was given as 80 square feet?
  2. How would the solution change if the length was 6 feet longer than the width?
  3. How would you solve this problem using factoring instead of the quadratic formula?
  4. What are the steps to derive the quadratic formula?
  5. How does the discriminant in the quadratic formula tell us about possible solutions?

Tip: When setting up quadratic equations for area problems, carefully identify which variable represents width and length to avoid misinterpretation.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Geometry

Formulas

Quadratic equation formula x^2 + bx + c = 0
Area of a rectangle (Area = length x width)

Theorems

Quadratic Formula

Suitable Grade Level

Grades 8-10