There are TWO solutions to a quadratic equation. Unless otherwise instructed, give both answers if it is a pure math problem. For real world problems and story problems, one of the answers is usually ridiculous and can be ignored. Do a sanity check to identify the applicable solution. In the case of astory problem, if you are unsure which solution is correct, give both. **Problem Statement:**For each of the following, set up either a system of equations in two variables or a quadratic equation and solve. (Round to two decimal places if necessary.)**10. Landscaping:**The perimeter of a rectangular field is 750 ft. If the length is four times the width, find the dimensions of the field.---**Solution Explanation:**1. **Define Variables:** - Let $ w $ be the width of the field. - Let $ l $ be the length of the field.2. **Equations Based on the Problem:** - The formula for the perimeter $ P $ of a rectangle is given by: \[ P = 2l + 2w \] - According to the problem, the length is four times the width: \[ l = 4w \] - The perimeter is given as 750 ft: \[ 2l + 2w = 750 \]3. **Substitute and Solve:** - Substitute $ l = 4w $ into the perimeter equation: \[ 2(4w) + 2w = 750 \] - Simplify and solve for $ w $: \[ 8w + 2w = 750 \] \[ 10w = 750 \] \[ w = 75 \]4. **Find Length:** - Use the relationship $ l = 4w $ to find $ l $: \[ l = 4(75) = 300 \]5. **Conclusion:** - The dimensions of the field are: - Width = 75 ft - Length = 300 ftThis structured approach allows you to solve for the dimensions of the field effectively using algebraic manipulation and substitution.

We know that the perimeter of a rectangle is 2(l+w) where l is length and w is width. We can set up…

For each of the following, set up either a system of equations in two variables or a quadratic equation and solve. (Round to two decimal places if necessary.) 10. Landscaping The perimeter of a rectangular field is 750 ft. If the length is four times the width, find the dimensions of the field.

For each of the following, set up either a system of equations in two variables or a quadratic equation and solve. (Round to two decimal places if necessary.) 10. Landscaping The perimeter of a rectangular field is 750 ft. If the length is four times the width, find the dimensions of the field.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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A: Introduction :

Question

There are TWO solutions to a quadratic equation. Unless otherwise instructed, give both answers if it is a pure math problem. For real world problems and story problems, one of the answers is usually ridiculous and can be ignored. Do a sanity check to identify the applicable solution. In the case of astory problem, if you are unsure which solution is correct, give both.

**Problem Statement:**

For each of the following, set up either a system of equations in two variables or a quadratic equation and solve. (Round to two decimal places if necessary.)

**10. Landscaping:**

The perimeter of a rectangular field is 750 ft. If the length is four times the width, find the dimensions of the field.

---

**Solution Explanation:**

1. **Define Variables:**
- Let $ w $ be the width of the field.
- Let $ l $ be the length of the field.

2. **Equations Based on the Problem:**
- The formula for the perimeter $ P $ of a rectangle is given by:
\[
P = 2l + 2w
\]
- According to the problem, the length is four times the width:
\[
l = 4w
\]
- The perimeter is given as 750 ft:
\[
2l + 2w = 750
\]

3. **Substitute and Solve:**
- Substitute $ l = 4w $ into the perimeter equation:
\[
2(4w) + 2w = 750
\]
- Simplify and solve for $ w $:
\[
8w + 2w = 750
\]
\[
10w = 750
\]
\[
w = 75
\]

4. **Find Length:**
- Use the relationship $ l = 4w $ to find $ l $:
\[
l = 4(75) = 300
\]

5. **Conclusion:**
- The dimensions of the field are:
- Width = 75 ft
- Length = 300 ft

This structured approach allows you to solve for the dimensions of the field effectively using algebraic manipulation and substitution.

Formula Formula A polynomial with degree 2 is called a quadratic polynomial. A quadratic equation can be simplified to the standard form: ax² + bx + c = 0 Where, a ≠ 0. A, b, c are coefficients. c is also called "constant". 'x' is the unknown quantity

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