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Ch. 1 - Equations and Inequalities
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 2, Problem 21

Dimensions of a Square. The length of each side of a square is 3 in. more than the length of each side of a smaller square. The sum of the areas of the squares is 149 in.2. Find the lengths of the sides of the two squares.

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1
Let the length of each side of the smaller square be represented by the variable \(x\) inches.
Since the larger square's side is 3 inches more than the smaller square's side, express the larger square's side length as \(x + 3\) inches.
Write expressions for the areas of both squares: the smaller square's area is \(x^2\) and the larger square's area is \((x + 3)^2\).
Set up an equation using the information that the sum of the areas is 149 square inches: \(x^2 + (x + 3)^2 = 149\).
Expand the squared term, simplify the equation, and solve the resulting quadratic equation for \(x\) to find the side lengths of the squares.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Algebraic Representation of Geometric Quantities

This involves translating geometric information into algebraic expressions. For example, representing the side lengths of squares as variables and expressing their areas as the square of those variables. This step is crucial for setting up equations based on the problem's conditions.
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Forming and Solving Quadratic Equations

When areas of squares are involved, the resulting equations often become quadratic. Understanding how to form a quadratic equation from the problem's conditions and solving it using factoring, completing the square, or the quadratic formula is essential to find the unknown side lengths.
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Interpreting and Verifying Solutions

After solving the quadratic equation, it's important to interpret the solutions in the context of the problem, ensuring they make sense (e.g., side lengths must be positive). Verifying solutions by substituting back into the original conditions confirms their correctness.
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