Solving a Garden’s Length and Width Using Quadratic Equations

Solving real-world problems using mathematics and algebra is both practical and rewarding. One such example involves determining the dimensions of a garden given its length and width relationship and area. Let's explore this problem step-by-step and see how mathematical concepts can provide us with the answers.

Understanding the Problem

The problem statement gives us the following information:

The width of the garden is represented by x meters. The length of the garden is x 5 meters. The area of the garden is 14 square meters.

Using these pieces of information, we can formulate a quadratic equation and solve for the width, from which we can easily determine the length.

Solving the Quadratic Equation

The area of the garden can be calculated by multiplying the length and width, which gives us the equation:

Length x Width Area

Substituting the known values, we get:

(x 5) * x 14

Expanding this equation, we obtain:

x^2 5x 14

Rearranging the equation into the standard quadratic form:

x^2 5x - 14 0

To solve this quadratic equation, we can use the quadratic formula:

x [-b ± sqrt(b^2 - 4ac)] / 2a

Here, a 1, b 5, and c -14. Substituting these values, we have:

x [-5 ± sqrt(5^2 - 4 * 1 * (-14))] / 2

Calculating the discriminant:

5^2 - 4 * 1 * (-14) 25 56 81

Plugging the discriminant back into the quadratic formula:

x [-5 ± sqrt(81)] / 2

Solving for x, we get two potential solutions:

x [-5 9] / 2 4 / 2 2 x [-5 - 9] / 2 -14 / 2 -7 (which is not a valid solution as width cannot be negative)

Therefore, the width of the garden is 2 meters. Substituting this value back into the equation for the length, we get:

Length 2 5 7 meters

Additional Examples for Practice

Let's explore additional examples to reinforce our understanding of quadratic equations in solving practical problems.

Example 1

Let the width of the garden be x meters. Then the length of the garden can be expressed as x 5 meters. The area of the garden is given by the formula:

Area Length * Width

Substituting the known values:

14 (x 5) * x

Expanding this equation:

14 x^2 5x

Rearranging it into the standard quadratic form:

x^2 5x - 14 0

Solving this quadratic equation using the quadratic formula, we get:

x [-5 ± sqrt(5^2 - 4 * 1 * (-14))] / 2

Calculating the discriminant:

5^2 - 4 * 1 * (-14) 25 56 81

Plugging the discriminant back into the quadratic formula:

x [-5 ± 9] / 2

Solving for x, we get two potential solutions:

x [4] / 2 2 x [-14] / 2 -7 (not valid as width cannot be negative)

Therefore, the width of the garden is 2 meters, and the length is 7 meters.

Example 2

Let the width of the garden be x feet. Then the length of the garden can be expressed as x 5 feet. The area of the garden is given by the formula:

Area Length * Width

Substituting the known values:

24 x * (x 5)

Expanding this equation:

24 x^2 5x

Rearranging it into the standard quadratic form:

x^2 5x - 24 0

Solving this quadratic equation using the quadratic formula, we get:

x [-5 ± sqrt(5^2 - 4 * 1 * (-24))] / 2

Calculating the discriminant:

5^2 - 4 * 1 * (-24) 25 96 121

Plugging the discriminant back into the quadratic formula:

x [-5 ± 11] / 2

Solving for x, we get two potential solutions:

x [6] / 2 3 x [-16] / 2 -8 (not valid as width cannot be negative)

Therefore, the width of the garden is 3 feet, and the length is 8 feet.

Conclusion

By understanding and applying quadratic equations, we can solve real-life problems related to dimensions and areas. This approach not only provides practical solutions but also enhances our mathematical problem-solving skills. For more resources, consider exploring additional examples and practice problems to further solidify your understanding of this mathematical concept.