Problem 269 - Elementary Algebra

Question

In the following exercises, simplify. (a) \(\quad \sqrt{11}(8+4 \sqrt{11})\) (b) \(\sqrt{3}(\sqrt{12}+\sqrt{27})\)

Step-by-Step Solution

Answer

a) 8√11 + 44 b) 15

Step 1: Simplify the Expression (a)

To simplify \(\sqrt{11}(8+4 \sqrt{11})\), distribute \(\sqrt{11}\) to each term inside the parentheses. This gives us \(\sqrt{11} \cdot 8 + \sqrt{11} \cdot 4 \sqrt{11}\).

Step 2: Distribute \(\sqrt{11}\)

Compute \( \sqrt{11} \cdot 8 + \sqrt{11} \cdot 4 \sqrt{11} \). This simplifies to \( 8\sqrt{11} + 4 \cdot \( \sqrt{11 \cdot 11} \)\).

Step 3: Simplify Radicals

Since \(\sqrt{11 \cdot 11} = \sqrt{11^2} = 11\), the expression becomes \(8\sqrt{11} + 4 \cdot 11\).

Step 4: Final Simplification

Combine the terms to get \( 8\sqrt{11} + 44 \). This is the simplified form of \(\sqrt{11}(8+4 \sqrt{11})\).

Step 5: Simplify the Expression (b)

To simplify \(\sqrt{3}(\sqrt{12}+\sqrt{27})\), start by simplifying \(\sqrt{12}\) and \(\sqrt{27}\).

Step 6: Simplify Radicals

Realize that \(\sqrt{12} = \sqrt{4\cdot3} = 2\sqrt{3}\) and \(\sqrt{27} = \sqrt{9\cdot3} = 3\sqrt{3}\). So, the expression becomes \(\sqrt{3}(2\sqrt{3}+3\sqrt{3})\).

Step 7: Combine Like Terms

Combine like terms inside the parentheses to get \(\sqrt{3}(5\sqrt{3})\).

Step 8: Multiply Radicals

Multiply the radicals: \(\sqrt{3} \cdot 5 \sqrt{3} = 5 \cdot (\sqrt{3 \cdot 3}) = 5 \cdot \sqrt{9} = 5 \cdot 3 = 15\). So, the simplified form of \(\sqrt{3}(\sqrt{12}+\sqrt{27})\) is 15.

Key Concepts

Simplifying Expressions

When you're simplifying expressions, the goal is to make a math statement as simple as possible. Each term and operation in an algebraic expression should be reduced to its simplest form. Look out for things like addition, subtraction, multiplication, and division.
Breaking down complex expressions step-by-step is crucial. In our exercise, we start with expressions involving square roots, also known as radicals.
Simplify one operation at a time, ensuring you manage and combine similar terms effectively. This approach helps us get rid of parentheses and combine like terms, making the expression shorter and easier to solve.

• Start with inside the parentheses.
• Remove grouping symbols by distributing or combining like terms.
• Keep your work organized to avoid mistakes.

Distributive Property

The distributive property is a useful algebraic tool that helps make expressions simpler and enables one to handle parentheses. The distributive property states that for any numbers or variables a, b, and c, the expression a(b + c) can be expanded to ab + ac.
In our example, we applied the distributive property to \(\backsqrt{11} (8 + 4 \backsqrt{11})\).\br> By distributing \(\backsqrt{11}\) to both terms inside the parentheses:\br \(\backsqrt{11} \cdot 8 + \backsqrt{11} \cdot 4 \backsqrt{11} \).\br This step breaks down our operations, removing the parentheses and setting us up for simplification.\br Remember, the distributive property works in reverse for factoring and can help you simplify expressions in operational order too.\br

• Distribute factors to each term inside a parenthesis.
• Perform operations step-by-step.
• Use the property to both expand and factor expressions.

Radicals

Radicals, often seen as square roots, involve expressing numbers in terms of their roots. When simplifying radicals, focus on prime factorization and combining like terms.
In calculations, handling radicals can seem complex. For example, consider the terms involving \(\backsqrt{12}\) and \(\backsqrt{27}\) in our exercise. We simplified these by expressing them in terms of their prime factors. \(\backsqrt{12} = \backsqrt{4 \cdot 3}\), knowing \(\backsqrt{4} = 2\), simplified to \(\backsqrt{12} = 2 \backsqrt{3}\).\br Likewise, \(\backsqrt{27} = \backsqrt{9 \cdot 3}\). Since \(\backsqrt{9} = 3\), it simplifies to \(\backsqrt{27} = 3 \backsqrt{3}\). Combining these like terms gives us simplified radicals.\br Remember, the simplified form of radicals reduces the complexity and helps combine terms effectively.\br

• Simplify radicals using prime factorization.
• Combine like terms to streamline expressions.
• Recognize properties of roots for easier simplification.