Chapter 7: Problem 48
Solve the given problems. All numbers are accurate to at least two significant digits. A homeowner wants to build a rectangular patio with an area of \(20.0 \mathrm{m}^{2},\) such that the length is \(2.0 \mathrm{m}\) more than the width. What should the dimensions be?
Short Answer
Expert verified
The dimensions of the patio are approximately 3.59 m by 5.59 m.
Step by step solution
01
Define Variables
Let the width of the patio be \( w \) meters. Since the length is \( 2.0 \;\mathrm{m} \) more than the width, we can express the length as \( w + 2.0 \;\mathrm{m} \).
02
Set Up the Area Equation
The area of a rectangle is given by multiplying its length by its width. Therefore, the equation for the area is \( (w)(w + 2.0) = 20.0 \;\mathrm{m}^{2} \).
03
Expand and Rearrange the Quadratic Equation
Expand the equation: \( w^2 + 2.0w = 20.0 \;\mathrm{m}^2 \). Rearrange it to form a standard quadratic equation: \( w^2 + 2.0w - 20.0 = 0 \).
04
Use the Quadratic Formula
The quadratic formula is \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For our equation, \( a = 1 \), \( b = 2.0 \), and \( c = -20.0 \). Substitute these values into the formula.
05
Calculate the Discriminant
Compute the discriminant \( b^2 - 4ac \): \( (2.0)^2 - 4(1)(-20.0) = 4 + 80 = 84 \).
06
Solve for Width \(w\)
Substitute the discriminant into the quadratic formula: \( w = \frac{-2.0 \pm \sqrt{84}}{2} \). Calculate \( \sqrt{84} \approx 9.17 \). Thus, \( w = \frac{-2.0 \pm 9.17}{2} \).
07
Determine Possible Solutions and Choose
Calculate the possible solutions: \( w = \frac{7.17}{2} \approx 3.585 \;\mathrm{m} \) and \( w = \frac{-11.17}{2} \) (not plausible as width can't be negative).
08
Find the Length from Width
Using the width \( w = 3.585 \;\mathrm{m} \), calculate the length: \( w + 2.0 = 3.585 + 2.0 = 5.585 \;\mathrm{m} \).
09
Verify the Calculation
Verify the area: \( 3.585 \times 5.585 \approx 20.0 \;\mathrm{m}^2 \), confirming our solution is correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area of a Rectangle
To understand the problem, we need to start with the basics of calculating the area of a rectangle. The area is the space inside the rectangle. It is calculated by multiplying the length by the width. So, the area formula is:
- Area = length × width
- Area = \( w \times (w + 2.0) \) = 20.0
Solving Quadratics
After setting up the equation \( w(w + 2.0) = 20.0 \), it becomes crucial to know how to solve quadratic equations. A quadratic equation typically takes the standard form:
- \( ax^2 + bx + c = 0 \)
- Combine like terms to simplify it as much as possible.
- Consider factoring as a method if it is an easy expression.
- If typical factoring does not work, use methods like completing the square or the quadratic formula.
Quadratic Formula
The quadratic formula is a powerful tool for finding the roots of any quadratic equation. This formula is written as:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- If positive, two distinct real solutions exist.
- If zero, exactly one real solution exists.
- If negative, real solutions do not exist (roots are complex).