A Rectangle's Length Is 2 Units More Than Twice Its Width. Its Area Is 40 Square Units. The Equation W ( 2 W + 2 ) = 40 W(2w+2) = 40 W ( 2 W + 2 ) = 40 Can Be Used To Find W W W , The Width Of The Rectangle.What Is The Width Of The Rectangle?A. 4 Units B. 5 Units C. 10

Introduction

In this article, we will delve into the world of mathematics and explore a problem involving a rectangle's dimensions. We will use algebraic equations to find the width of the rectangle, given its area and the relationship between its length and width. This problem is a great example of how mathematical concepts can be applied to real-world scenarios.

The Problem

A rectangle's length is 2 units more than twice its width. Its area is 40 square units. We can use the equation $w(2w+2) = 40$ to find $w$, the width of the rectangle.

Understanding the Equation

The equation $w(2w+2) = 40$ represents the relationship between the width and the area of the rectangle. To solve for $w$, we need to isolate the variable $w$ on one side of the equation.

Expanding the Equation

We can start by expanding the equation using the distributive property:

$$w(2w+2) = 40$$

$$2w^2 + 2w = 40$$

Rearranging the Equation

Next, we can rearrange the equation to get all the terms on one side:

$$2w^2 + 2w - 40 = 0$$

Solving the Quadratic Equation

The equation $2w^2 + 2w - 40 = 0$ is a quadratic equation, which can be solved using various methods. We can use the quadratic formula to find the solutions:

$$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 2$, $b = 2$, and $c = -40$. Plugging these values into the formula, we get:

$$w = \frac{-2 \pm \sqrt{2^2 - 4(2)(-40)}}{2(2)}$$

$$w = \frac{-2 \pm \sqrt{4 + 320}}{4}$$

$$w = \frac{-2 \pm \sqrt{324}}{4}$$

$$w = \frac{-2 \pm 18}{4}$$

Finding the Solutions

We now have two possible solutions for $w$:

$$w = \frac{-2 + 18}{4} = \frac{16}{4} = 4$$

$$w = \frac{-2 - 18}{4} = \frac{-20}{4} = -5$$

However, since the width of a rectangle cannot be negative, we discard the solution $w = -5$.

Conclusion

The width of the rectangle is $\boxed{4}$ units.

Discussion

This problem is a great example of how mathematical concepts can be applied to real-world scenarios. The use of algebraic equations to find the width of the rectangle demonstrates the importance of mathematical problem-solving skills. The quadratic formula is a powerful tool for solving quadratic equations, and its application in this problem highlights its usefulness.

Related Topics

• Quadratic equations
• Algebraic equations
• Mathematical problem-solving
• Rectangle dimensions

Further Reading

For more information on quadratic equations and algebraic equations, please refer to the following resources:

• Quadratic Equation Formula
• Algebraic Equations
• Rectangle Dimensions
  A Rectangle's Width: A Mathematical Exploration - Q&A ===========================================================

Introduction

In our previous article, we explored the problem of finding the width of a rectangle given its area and the relationship between its length and width. We used algebraic equations to find the width of the rectangle, and in this article, we will answer some frequently asked questions related to this problem.

Q&A

Q: What is the relationship between the length and width of the rectangle?

A: The length of the rectangle is 2 units more than twice its width. This can be represented by the equation $l = 2w + 2$, where $l$ is the length and $w$ is the width.

Q: How do we find the width of the rectangle?

A: We can use the equation $w(2w+2) = 40$ to find the width of the rectangle. This equation represents the relationship between the width and the area of the rectangle.

Q: What is the quadratic formula, and how is it used to solve quadratic equations?

A: The quadratic formula is a mathematical formula that is used to solve quadratic equations. It is given by the equation $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation.

Q: Why do we discard the negative solution for the width of the rectangle?

A: We discard the negative solution for the width of the rectangle because the width of a rectangle cannot be negative. A negative width would not make sense in the context of the problem.

Q: What are some real-world applications of quadratic equations?

A: Quadratic equations have many real-world applications, including:

• Projectile motion: Quadratic equations can be used to model the trajectory of a projectile, such as a thrown ball or a rocket.
• Optimization: Quadratic equations can be used to optimize functions, such as finding the maximum or minimum value of a function.
• Physics: Quadratic equations can be used to model the motion of objects, such as the motion of a pendulum or a spring.

Q: How can I practice solving quadratic equations?

A: There are many resources available to help you practice solving quadratic equations, including:

• Online practice problems: Websites such as Khan Academy and Mathway offer online practice problems to help you practice solving quadratic equations.
• Textbooks: Many algebra textbooks include practice problems and exercises to help you practice solving quadratic equations.
• Math apps: There are many math apps available that can help you practice solving quadratic equations, such as Photomath and Mathway.

Conclusion

In this article, we have answered some frequently asked questions related to the problem of finding the width of a rectangle given its area and the relationship between its length and width. We have also discussed the quadratic formula and its application in solving quadratic equations. We hope that this article has been helpful in clarifying any questions you may have had about this problem.

Related Topics

• Quadratic equations
• Algebraic equations
• Mathematical problem-solving
• Rectangle dimensions

Further Reading

For more information on quadratic equations and algebraic equations, please refer to the following resources:

• Quadratic Equation Formula
• Algebraic Equations
• Rectangle Dimensions