Math Problem Statement

Can you solve this problem using this formula L x W and use factoring

The length of a rectangular backyard is 9 meters more than the width. What are the dimensions of the garden that will make the area equal to 136 m²?

Solution

Let's solve this problem step by step, using the given formula L×WL \times W (Length times Width), and apply factoring.

Problem Restatement:

We are told:

  • The length of a rectangular backyard is 9 meters more than its width.
  • The area of the backyard is 136 square meters.

Let:

  • WW be the width of the backyard.
  • LL be the length of the backyard.

From the information, we know that:

  • L=W+9L = W + 9 (since the length is 9 meters more than the width).
  • The area AA is given by the formula L×W=136L \times W = 136.

Step 1: Set up the equation using the area formula

Substitute L=W+9L = W + 9 into the area formula:

(W+9)×W=136(W + 9) \times W = 136

Step 2: Expand the equation

Expand the left-hand side:

W2+9W=136W^2 + 9W = 136

Step 3: Rearrange into standard quadratic form

Move all terms to one side to set the equation to 0:

W2+9W136=0W^2 + 9W - 136 = 0

Step 4: Factor the quadratic equation

We need to factor W2+9W136=0W^2 + 9W - 136 = 0.

We are looking for two numbers that multiply to give 136-136 and add up to 99. The correct pair is 1717 and 8-8 because:

17×8=136and17+(8)=917 \times -8 = -136 \quad \text{and} \quad 17 + (-8) = 9

Thus, the quadratic equation factors as:

(W+17)(W8)=0(W + 17)(W - 8) = 0

Step 5: Solve for WW

Set each factor equal to 0:

W+17=0orW8=0W + 17 = 0 \quad \text{or} \quad W - 8 = 0

Solving these gives:

W=17orW=8W = -17 \quad \text{or} \quad W = 8

Since the width cannot be negative, we take W=8W = 8.

Step 6: Solve for LL

Substitute W=8W = 8 back into L=W+9L = W + 9:

L=8+9=17L = 8 + 9 = 17

Final Answer:

The dimensions of the backyard are:

  • Width W=8W = 8 meters.
  • Length L=17L = 17 meters.

Conclusion:

The dimensions that make the area equal to 136 square meters are 8 meters (width) and 17 meters (length).

Would you like more details or have any questions?


Related Questions:

  1. How would the dimensions change if the length were 12 meters more than the width?
  2. Can the quadratic equation W2+9W136=0W^2 + 9W - 136 = 0 be solved using the quadratic formula?
  3. What would the area be if the width was doubled?
  4. How can you confirm the solution by recalculating the area with the dimensions found?
  5. How does factoring compare to solving quadratic equations by completing the square?

Tip: Factoring is a fast and efficient method to solve quadratic equations, but it only works when the quadratic can be factored neatly into integer values.

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Quadratic Equations

Formulas

Area of a rectangle: L × W
Quadratic equation: ax² + bx + c = 0

Theorems

Factoring
Zero Product Property

Suitable Grade Level

Grades 8-10