Example

Finding the Dimensions of a Rectangular Garden with Area 15 Square Feet

Apply the seven-step geometry problem-solving strategy to a rectangle area problem where one dimension is expressed in terms of the other, producing a quadratic equation that must be solved by factoring.

Problem: A rectangular garden has an area of 1515 square feet. The length of the garden is two feet more than the width. Find the length and width of the garden.

  1. Read: A rectangular garden has A=15A = 15 sq ft, and its length is 22 feet more than its width. Draw and label the rectangle with width WW and length W+2W + 2.
  2. Identify: The length and width of the garden.
  3. Name: Let WW = the width of the garden. Then W+2W + 2 = the length of the garden.
  4. Translate: Write the area formula and substitute:

A=LWA = L \cdot W

15=(W+2)W15 = (W + 2)W

  1. Solve: Distribute: 15=W2+2W15 = W^2 + 2W. Subtract 1515 from both sides to obtain standard form: 0=W2+2W150 = W^2 + 2W - 15. Factor the trinomial by finding two numbers whose product is 15-15 and whose sum is 22: the pair 55 and 3-3 works. So 0=(W+5)(W3)0 = (W + 5)(W - 3). Apply the Zero Product Property: W+5=0W + 5 = 0 or W3=0W - 3 = 0, giving W=5W = -5 or W=3W = 3. Since width cannot be negative, discard W=5W = -5. Therefore W=3W = 3. The length is 3+2=53 + 2 = 5.
  2. Check: Area = 3×5=153 \times 5 = 15 sq ft ✓. The length 55 is indeed two more than the width 33 ✓.
  3. Answer: The width of the garden is 33 feet and the length is 55 feet.

Unlike earlier rectangle problems that used the perimeter formula and produced linear equations, this problem uses the area formula with one dimension expressed relative to the other. Substituting the expression W+2W + 2 for the length into A=LWA = L \cdot W creates the product (W+2)W(W + 2)W, which expands into a quadratic expression W2+2WW^2 + 2W. After rearranging to standard form, the equation is solved by factoring and applying the Zero Product Property. The negative solution W=5W = -5 is algebraically valid but must be discarded because a physical width cannot be negative.

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Updated 2026-04-21

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