Question

Dimensions of a Window The area of the opening of a rectangular window is to be 143 square feet. If the length is to be 2 feet more than the width, what are the dimensions?

Short answer

Width: 11 feet, Length: 13 feet.

Step-by-step solution

Define Variables

Let the width of the window be denoted as \(w\) feet. Since the length is 2 feet more than the width, the length can be written as \(w + 2\) feet.

Write the Area Equation

We know that the area of a rectangle is given by the formula \( \text{Area} = \text{Length} \times \text{Width} \). Substituting the given values, we get the equation: \[ \text{Area} = (w + 2) \times w = 143 \]

Solve the Quadratic Equation

Expand and simplify the quadratic equation:\[ w^2 + 2w - 143 = 0 \]Now, solve for \(w\) using the quadratic formula \( w = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 2 \), and \( c = -143 \).

Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \):\[ 2^2 - 4(1)(-143) = 4 + 572 = 576 \]

Find the Width

Substitute the discriminant back into the quadratic formula:\[ w = \frac{-2 \, \pm \, \sqrt{576}}{2} = \frac{-2 \, \pm \, 24}{2} \]This yields two solutions:\[ w = \frac{22}{2} = 11 \text{ feet} \] and \[ w = \frac{-26}{2} = -13 \text{ feet} \]Since width cannot be negative, we have \( w = 11 \text{ feet} \).

Find the Length

The length is \( w + 2 \), so substituting \( w = 11 \):\[ \text{Length} = 11 + 2 = 13 \text{ feet} \]

Key concepts

solving quadratic equations

Quadratic equations are fundamental in algebra. They are equations of the form \(ax^2 + bx + c = 0\), where \(a e 0\). Solving these equations means finding the values of \(x\) that make the equation true.
To solve a quadratic equation, we often use the quadratic formula: \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \). Here, the solutions \(x\) are found by substituting the values of \(a\), \(b\), and \(c\) from the quadratic equation. The plus-minus (\(\pm\)) signs indicate that there are generally two solutions.
Another method to solve quadratic equations is factoring when possible or completing the square.
Practice solving quadratic equations with different values of \(a\), \(b\), and \(c\) to become comfortable with the process. Remember, the key is to isolate the variable \(x\) and solve the equation step by step.

area of rectangles

Understanding the area of rectangles is crucial for solving many geometry problems. A rectangle’s area is calculated using the formula: \( \text{Area} = \text{Length} \times \text{Width} \).
In problems like the one provided, knowing the area and one dimension allows us to find the other dimension by setting up an equation.
Given the width \(w\) and the length as \(w + 2\), we substitute these into the area formula and solve for \(w\).
In more general terms:

• The area is always a product of two sides (length and width).
• Units are typically given in square feet, square meters, etc.
• Given the area and one side, you can always rearrange the formula to find the missing side.

quadratic formula

The quadratic formula is an essential tool for solving quadratic equations when they cannot be easily factored. Here is the formula again: \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \).
This formula works for any quadratic equation, provided you correctly identify the coefficients \(a\), \(b\), and \(c\).
Steps to use the quadratic formula:

• Write down the quadratic equation and identify \(a\), \(b\), and \(c\).
• Calculate the discriminant: \( b^2 - 4ac \).
• Substitute \(a\), \(b\), \(c\), and the discriminant into the formula.
• Simplify to find the solutions for \(x\).

Remember, the quadratic formula reveals the solutions even when they are complex or irrational.
Practice different equations to see how versatile this formula is.

discriminant

The discriminant is a component of the quadratic formula that gives insight into the nature of the solutions. It is calculated as \( b^2 - 4ac \).
The value of the discriminant determines the type of solutions a quadratic equation will have:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (also known as a repeated root).
• If the discriminant is negative, there are two complex solutions (no real solutions).

The discriminant lets us know what kind of numbers to expect before solving the equation.
In our exercise, the discriminant was positive (576), which led to two real solutions for the width. However, only one solution made sense in the context of the problem (positive width).
Understanding discriminants is crucial for predicting and verifying the nature of solutions to quadratic equations.