Problem 35 - Algebra 2 and Trigonometry

Question

The length of a rectangle is 4 feet more than twice the width. The area of the rectangle is 38 square feet. a. Find the dimensions of the rectangle in simplest radical form. b. Show that the product of the length and width is equal to the area. c. Write, to the nearest tenth, rational approximations for the length and width.

Step-by-Step Solution

Answer

The width is \(-1 + 2\sqrt{5}\) feet, and the length is \(4\sqrt{5} + 2\) feet. The rational approximations are 3.4 feet for the width and 10.8 feet for the length.

Step 1: Identify Variables and Equations

Let the width of the rectangle be denoted by \( w \). According to the problem, the length \( l \) is 4 feet more than twice the width, so we can express this as \( l = 2w + 4 \). The area of the rectangle is given by \( A = l \times w = 38 \). Thus, we have the equation \((2w + 4)w = 38\).

Step 2: Formulate the Quadratic Equation

Expand the equation \((2w + 4)w = 38\) to get \(2w^2 + 4w = 38\). Rearranging terms gives us the quadratic equation \(2w^2 + 4w - 38 = 0\).

Step 3: Solve the Quadratic Equation

Use the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = 4\), and \(c = -38\). Substitute these values into the formula to get \(w = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 2 \cdot (-38)}}{2 \cdot 2}\). Simplify to find \(w = \frac{-4 \pm \sqrt{16 + 304}}{4}\), which reduces to \(w = \frac{-4 \pm \sqrt{320}}{4}\). Since \(320 = 64 \times 5 = 8^2 \times 5\), \(\sqrt{320} = 8\sqrt{5}\). Hence, \(w = \frac{-4 \pm 8\sqrt{5}}{4}\).

Step 4: Simplify the Radical Expression

From \(w = \frac{-4 \pm 8\sqrt{5}}{4}\), simplify \(w\) by dividing each term: \(w = -1 \pm 2\sqrt{5}\). Since width cannot be negative, choose \(w = -1 + 2\sqrt{5}\).

Step 5: Find the Length

Substitute \(w = -1 + 2\sqrt{5}\) into the expression for length: \(l = 2w + 4\). Therefore, \(l = 2(-1 + 2\sqrt{5}) + 4 = -2 + 4\sqrt{5} + 4 = 4\sqrt{5} + 2\).

Step 6: Verify the Area

Verify that the product \(l \times w\) equals the area by calculating \((4\sqrt{5} + 2)(-1 + 2\sqrt{5})\). Simplifying this product should equal 38 square feet, confirming our calculations.

Step 7: Rational Approximation of Width and Length

To find rational approximations, approximate \(\sqrt{5} \approx 2.2\). Hence, \(w = -1 + 2 \times 2.2 = 3.4\) and \(l = 4 \times 2.2 + 2 = 10.8\). These are the dimensions to the nearest tenth.

Key Concepts

Simplifying Radical Expressions

When working with quadratic equations, you often encounter square roots, leading to expressions involving radicals. Simplifying radical expressions is a crucial step to make the results more manageable and understandable.
To simplify such expressions, look for perfect squares within the radicand (the number under the square root). For instance, if you encounter \(\sqrt{320}\), you can express 320 as the product of 64 and 5, where 64 is a perfect square. Thus, \(\sqrt{320} = \sqrt{64 \times 5} = \sqrt{64} \times \sqrt{5}\).
Knowing that \(\sqrt{64} = 8\), you simplify \(\sqrt{320}\) to \(8\sqrt{5}\).

• Factor the radicand to identify perfect squares.
• Take the square root of the perfect squares.
• Multiply the results to simplify further.

Understanding this process is vital because it simplifies calculations in algebraic operations, making it easier to find exact solutions.

Algebraic Expressions

Algebraic expressions form the backbone of working with equations in algebra. They entail numbers, variables, and operations, knitted together by algebraic rules to represent a value or a set of values.
In the context of this problem, identifying the lengths and widths of a rectangle requires forming an algebraic expression. Given the width \(w\), the length \(l\) is formulated as \(l = 2w + 4\) since it is 4 feet more than twice the width.
Writing and understanding these expressions helps establish relationships between the variables and allows us to set up equations that can be solved to find unknown values.

• Variables represent unknown values; in this problem, the width is denoted by \(w\).
• Use given conditions to express one variable in terms of another.
• Apply algebraic operations consistently to manipulate and solve these expressions.

This systematic approach aids in simplifying complex problems, allowing for straightforward solutions.

Solving Quadratic Equations

Quadratic equations are a staple in algebra, often appearing in scenarios involving areas, projectile motions, and optimization problems. They have the general form \(ax^2 + bx + c = 0\). Solving these equations typically involves finding the values of \(x\) that satisfy the equation.
In our exercise, we derived the equation \(2w^2 + 4w - 38 = 0\) from the product of the rectangle's dimensions equating to its area. This is a typical quadratic equation. To solve it, the quadratic formula \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) comes into play.
Substituting \(a = 2\), \(b = 4\), and \(c = -38\) into the formula allows us to find the potential widths. One viable solution arises when focusing on the positive result of the radical, leading to \(w = -1 + 2\sqrt{5}\).

• Recognize the standard form: \(ax^2 + bx + c = 0\).
• Use the quadratic formula for solving.
• Ensure the solution fits within the constraints of the problem, like positive dimensions.

Understanding how to manipulate and solve quadratic equations is pivotal in addressing real-world problems mathematically.