Problem 20 - Algebra 2 and Trigonometry

Question

In \(19-25,\) express each answer in simplest radical form. Check each answer. The length of a rectangle is 2 feet more than the width. The area of the rectangle is 2 square feet. What are the dimensions of the rectangle?

Step-by-Step Solution

Answer

The dimensions are \( (-1 + \sqrt{3}) \) feet by \( (1 + \sqrt{3}) \) feet.

Step 1: Define Variables

Let the width of the rectangle be \( w \) (in feet). Then, the length of the rectangle, which is 2 feet more than the width, will be \( w + 2 \).

Step 2: Write the Area Equation

The area of the rectangle is given by the formula \( \text{Area} = \text{length} \times \text{width} \). In terms of our variables, the equation becomes \( (w)(w + 2) = 2 \).

Step 3: Set Up the Quadratic Equation

Expand and rearrange the equation from Step 2: \( w^2 + 2w = 2 \). Bring all terms to one side to form a quadratic equation: \( w^2 + 2w - 2 = 0 \).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -2 \). Substitute these values in: \( w = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-2)}}{2 \times 1} \).

Step 5: Simplify and Solve for w

Calculate the discriminant: \( 2^2 - 4 \times 1 \times (-2) = 4 + 8 = 12 \). Therefore, \( w = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} \). Simplify: \( w = -1 \pm \sqrt{3} \).

Step 6: Determine Valid Width

Since a width cannot be negative, choose \( w = -1 + \sqrt{3} \).

Step 7: Calculate the Length

The length is \( w + 2 = (-1 + \sqrt{3}) + 2 = 1 + \sqrt{3} \).

Step 8: Check the Solution

Check if the area is correct: \( \text{Area} = (-1 + \sqrt{3})(1 + \sqrt{3}) = (-1)(1) + (-1)(\sqrt{3}) + (\sqrt{3})(1) + (\sqrt{3})(\sqrt{3}) = -1 - \sqrt{3} + \sqrt{3} + 3 \). Simplifying gives \( 2 \), confirming our solution.

Key Concepts

Radical Expressions

Radical expressions are mathematical expressions that include a root symbol, such as a square root \( \sqrt{} \). When solving quadratic equations, particularly those resulting in non-perfect squares under the root, radical expressions come into play. For instance, in this exercise, we found that the width of the rectangle is \(-1 + \sqrt{3} \), which is expressed in its simplest radical form.

Simplifying radical expressions involves rewriting the root in its simplest form. This could mean factoring out squares. The aim is to write the expression in such a way that any radicals have been simplified as far as possible.

In our exercise, we had to compute \( \sqrt{12} \) which simplifies to \( 2\sqrt{3} \), because we can factor 12 into \( 4 \times 3 \), where \( 4 \) is a perfect square.

• Remember that similar terms under the radical can often be combined or simplified.
• Always express your final answer in the simplest radical form.

Rectangle Dimensions

Understanding rectangle dimensions is key to setting up your problem correctly. In this exercise, we needed to determine both the width and the length of a rectangle where the length is 2 feet more than the width.

We began by defining variables:

• Let the width be denoted as \( w \).
• Thus, the length is \( w + 2 \).

This approach allows you to express both dimensions in terms of one variable, aiding in forming a quadratic equation that represents the area condition given in the prompt.

By doing this, we've put ourselves in a position to solve for one unknown (\( w \)), which then automatically gives us the length once \( w \) is determined. This method simplifies the problem, making it easier to solve.

Quadratic Formula

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). In the rectangle problem, our equation was \( w^2 + 2w - 2 = 0 \). We solved this using the quadratic formula, which is a reliable method to find solutions to any quadratic equation.

The quadratic formula is:
\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• The coefficients \( a \), \( b \), and \( c \) are taken from the equation.
• The term under the square root \( (b^2 - 4ac) \) is called the discriminant.
• If the discriminant is positive, you'll have two real solutions.

This formula helped us find the width of the rectangle by substituting \( a = 1 \), \( b = 2 \), and \( c = -2 \). The discriminant was 12, giving two potential solutions for \( w \), out of which we selected the only realistic value in the context of physical dimensions.

Area of Rectangle

The area of a rectangle is calculated using the formula \( \text{Area} = \text{length} \times \text{width} \). This formula forms the basis of many practical problems involving rectangles, including this exercise.

Setting up the correct equation for the area is crucial for solving problems related to dimensions and area.

• For our problem, the area was given as 2 square feet.
• We expressed the area equation using our variables: \( (w)(w + 2) = 2 \).

With this equation, we expanded and transformed it into a quadratic form. Solving this gives the values for the dimensions, which need to be checked for validity by plugging back into the area formula.

Always ensure the dimensions you find satisfy the original area condition. Confirm each calculation, as done in this exercise, to verify that the resulting area indeed matches the given, in this case, confirming that \( 2 \) square feet is correct.