Problem 230 - Elementary Algebra

Question

Suzy wants to use square tiles on the border of a spa she is installing in her backyard. She will use large tiles that have area of 12 square inches, medium tiles that have area of 8 square inches, and small tiles that have area of 4 square inches. Once section of the border will require 4 large tiles, 8 medium tiles, and 10 small tiles to cover the width of the wall. Simplify the expression \(4 \sqrt{12}+8 \sqrt{8}+10 \sqrt{4}\) to determine the width of the wall. (Round to the nearest tenth of an inch.)

Step-by-Step Solution

Answer

The width of the wall is approximately 56 inches.

Step 1 - Simplify Individual Square Roots

Begin by simplifying the square root of each tile's area. For large tiles, it is \( \sqrt{12} = 2 \sqrt{3} \). For medium tiles, it is \( \sqrt{8} = 2 \sqrt{2} \). For small tiles, it is \( \sqrt{4} = 2 \).

Step 2 - Substitute the Simplified Roots

Substitute the simplified square roots into the original expression. So, \( 4 \sqrt{12} + 8 \sqrt{8} + 10 \sqrt{4} \) turns into \( 4(2 \sqrt{3}) + 8(2 \sqrt{2}) + 10(2) \).

Step 3 - Apply Distributive Property

Use the distributive property to simplify further: \( 4 \times 2 \sqrt{3} = 8 \sqrt{3} \), \( 8 \times 2 \sqrt{2} = 16 \sqrt{2} \), and \( 10 \times 2 = 20 \).

Step 4 - Combine All Terms

Combine all the terms together to get the final simplified expression: \( 8 \sqrt{3} + 16 \sqrt{2} + 20 \).

Step 5 - Approximate Values

Approximate the values of the square roots, where \( \sqrt{3} \approx 1.7 \) and \( \sqrt{2} \approx 1.4 \).

Step 6 - Calculate Approximations

Calculate the products: \( 8 \times 1.7 \approx 13.6 \) and \( 16 \times 1.4 \approx 22.4 \), then add the constants: \( 13.6 + 22.4 + 20 \).

Step 7 - Sum Up the Results

Sum up all the approximated values: \( 13.6 + 22.4 + 20 = 56 \. \)

Key Concepts

Square Roots

Square roots are a fundamental concept in solving many mathematical problems, especially when dealing with areas and length measurements. A square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 4 is 2 because 2 \times 2 = 4. In Suzy's tile problem, understanding square roots is key to simplifying the expression for the width of the wall. For large tiles, \(\sqrt{12}\) simplifies to \(2\sqrt{3}\), for medium tiles, \(\sqrt{8}\) simplifies to \(2\sqrt{2}\), and for small tiles, \(\sqrt{4}\) simplifies to 2.

Breaking down the square roots allows us to handle more complex expressions systematically. It's essential to become comfortable with simplifying square roots, as this skill is often used in various fields like geometry, physics, and engineering.

Distributive Property

The distributive property is one of the most powerful tools in algebra and helps simplify expressions involving multiplication and addition. It states that for any numbers a, b, and c, the equation \(a(b + c) = ab + ac\) holds true. In Suzy's problem, after simplifying the square roots, we need to distribute them across the coefficients. For instance, \(4\sqrt{12} + 8\sqrt{8} + 10\sqrt{4}\) simplifies to \(4(2\sqrt{3}) + 8(2\sqrt{2}) + 10(2)\).

This step-by-step approach makes complex equations easier to manage. Using the distributive property lets us break down larger problems into smaller, more understandable pieces and leads us closer to the solution efficiently. Applying the distributive property helps achieve the intermediate result: \(8\sqrt{3} + 16\sqrt{2} + 20\).

Radical Expressions

A radical expression involves roots, such as square roots or cube roots. They are common in geometry and algebra. In Suzy's case, the original expression \(4\sqrt{12} + 8\sqrt{8} + 10\sqrt{4}\) is a radical expression that requires simplification for practical use, such as measuring the width of a wall. Simplifying the square roots first, and then using algebraic tools like the distributive property, helps resolve such expressions.

When dealing with radical expressions, it's essential to know how to combine like terms, simplify roots, and use algebraic properties. This approach makes working with expressions involving radicals straightforward and manageable.

Approximations

Approximations are often necessary in real-world applications when exact values are either impossible or impractical to use. In Suzy's task, we approximate \(\sqrt{3} \) and \(\sqrt{2} \) to decimal values for ease of calculation. Specifically, \(\sqrt{3} \approx 1.7\) and \(\sqrt{2} \approx 1.4\). Using these approximations lets us convert the expression \(8\sqrt{3} + 16\sqrt{2} + 20\) into \(8(1.7) + 16(1.4) + 20\).

Calculating further, this gives us: \(8 \times 1.7 = 13.6\) and \(16 \times 1.4 = 22.4\). Summing these values with the constant: \(13.6 + 22.4 + 20 = 56\), reveals that the width of the wall is approximately 56 inches. Approximations help make complex problems more manageable and provide practical solutions in everyday scenarios.