Problem 148 - College Algebra

Question

Each side of a square is lengthencd by 2 inches. The area of this new, larger square is 36 square inches. Find the length of a side of the original square.

Step-by-Step Solution

Answer

The length of a side of the original square is 4 inches.

Step 1: Identify the Formula for the Area of a Square

The formula for the area of a square is \(SideLength^2\). In this problem, the side length of the new square is the original side length plus 2 inches.

Step 2: Set Up the Equation

Given that the area of the larger square is 36 square inches, it can be expressed in terms of the original side length. Let \(a\) be the original side length, thus the equation is \((a + 2)^2=36\)

Step 3: Solve for the Original Side Length

Solve the equation from step 2 for \(a\). Expand out \((a + 2)^2\) to get \(a^2 + 4*a + 4 = 36\). Subtract 36 from both sides to set the equation to zero: \(a^2 + 4*a - 32 = 0\). This is a quadratic equation that can be factored into \((a - 4)(a + 8) = 0\). Setting each factor equal to zero gives the solutions \(a = 4\) and \(a = -8\). Since a length cannot be negative, the original side length is 4 inches.

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that typically take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. The solutions to these equations are the values of \(x\) that make the equation true. They are fundamental in algebra and appear in various contexts, including geometry, physics, and economics.

When solving quadratic equations, we often aim to find the roots of the equation, which are the values of \(x\) where the function crosses the \(x\)-axis. These roots can be found in different ways, including factoring, completing the square, using the quadratic formula, or graphing. The quadratic equation we encounter in our exercise is \(a^2 + 4a - 32 = 0\), which can be solved by factoring or applying the quadratic formula.

Area of a Square

The area of a square, a shape with four equal sides and right angles, is found by squaring the length of one of its sides. The formula for calculating the area is \(Area = SideLength^2\). This is because you're essentially multiplying the length of a side by itself since a square has equal sides. Understanding this concept is crucial in many geometry problems.

In our original problem, we're given the area of a square and instructed to find the side's length. This involves reversing the area formula process, which can lead to the formation of a quadratic equation if the side's length is not known directly, as seen in the exercise where the side length was altered by adding 2 inches.

Algebraic Problem Solving

Algebraic problem solving is the process of finding unknown values by applying mathematical principles and operations. It often involves setting up an equation based on the problem's conditions and then manipulating this equation to solve for the unknown.

In the context of the exercise, we have initially set up an equation based on the given area of the new, larger square. The equation consists of an expression, \(a + 2\), which accounts for the increased side length, being squared. By systematically expanding and simplifying this expression, and then solving the resulting quadratic equation, the value of the original side length 'a' can be determined. This kind of systematic approach is essential for breaking down and solving algebraic problems.

Factoring Quadratic Equations

Factoring quadratic equations is a method used to solve them by expressing the equation as a product of its factors. It works on the principle that if a product equals zero, at least one of the multiplicands must equal zero. To factor a quadratic equation, look for two numbers that multiply to give the constant term (the \(c\) in \(ax^2 + bx + c = 0\)) and add to give the coefficient of the \(x\) term (the \(b\)).

In our given exercise, the quadratic equation \(a^2 + 4a - 32 = 0\) was factored into \(a - 4)(a + 8) = 0\). Each factor is then set to zero, yielding potential solutions for \(a\). However, since the context of the problem involves finding a length, the negative solution is discarded, leaving the positive solution as the length of the side of the original square. This highlights the importance of considering the context and constraints of a problem when interpreting the solutions obtained through algebra.