Problem 59

Question

Multiply. Then simplify if possible. Assume that all variables represent positive real numbers. $$ \sqrt{2}(\sqrt{2}+x \sqrt{6}) $$

Step-by-Step Solution

Verified
Answer
The expression simplifies to \(2 + 2x \sqrt{3}\).
1Step 1: Distribute \(\sqrt{2}\) to Each Term
Distribute the \(\sqrt{2}\) to each term inside the parentheses: \((\sqrt{2})(\sqrt{2}) + (\sqrt{2})(x \sqrt{6})\).
2Step 2: Simplify Each Term
Simplify each resulting term. First, \((\sqrt{2})(\sqrt{2}) = 2\), since \(\sqrt{a} \cdot \sqrt{a} = a\). Next, \((\sqrt{2})(x \sqrt{6}) = x \sqrt{12}\) because \((a^m \cdot b^n) = (ab)^{(m+n)/2}\).
3Step 3: Simplify \(x \sqrt{12}\)
Simplify \(x \sqrt{12}\) as follows: \(x \sqrt{12} = x \sqrt{4 \cdot 3} = x \sqrt{4} \cdot \sqrt{3} = x \cdot 2 \cdot \sqrt{3} = 2x \sqrt{3}\).
4Step 4: Combine the Simplified Terms
Combine the simplified terms to write the final expression: \(2 + 2x \sqrt{3}\).

Key Concepts

Multiplication of RadicalsDistributive PropertySimplifying Radicals
Multiplication of Radicals
Multiplying radicals might seem tricky at first, but it's actually a straightforward process once you understand how to handle them. Radicals, or roots, can be treated similarly to regular numbers when multiplying.
Here's what to keep in mind:
  • When you multiply two radicals with the same root, you can combine them under a single radical. For example, \( \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \).
  • Radicals of the same index can be multiplied by multiplying the numbers inside the radicals.
In our specific exercise, we apply the multiplication of radicals when multiplying \( \sqrt{2} \) by each term inside the parentheses. This means we have to multiply \( \sqrt{2} \) by both \( \sqrt{2} \) and \( x \sqrt{6} \). Let's break it down:
  • First, \( \sqrt{2} \cdot \sqrt{2} \) is simply 2. This is because \( \sqrt{2} \cdot \sqrt{2} = \sqrt{4} = 2 \).
  • Then, \( \sqrt{2} \cdot x \sqrt{6} \) gives us \( x \sqrt{12} \). This happens because we're multiplying the numbers under the radicals (2 and 6) to get \( \sqrt{12} \).
Once you understand these principles, multiplying radicals becomes just another part of solving algebraic expressions!
Distributive Property
The distributive property is a critical concept in algebra, acting almost like a bridge for combining different components of an expression. It states that \( a(b + c) = ab + ac \). This means you multiply each term inside the parentheses by the factor outside.
In our example, we're using the distributive property to handle the expression \( \sqrt{2} (\sqrt{2} + x\sqrt{6}) \).
  • First, distribute \( \sqrt{2} \) to \( \sqrt{2} \).
  • Next, distribute \( \sqrt{2} \) to \( x \sqrt{6} \).
The distributive property helps us break down expressions into parts that can be simplified individually. This step is vital because it sets up the equation so that we can easily apply further simplifications, like multiplying and simplifying radicals. Understanding and mastering the distributive property empowers students to manipulate and solve complex algebraic equations with confidence.
Simplifying Radicals
Simplifying radicals is an essential skill for making algebraic expressions more manageable. Simplification involves reducing the expression to its simplest form, often making the numbers easier to work with.
Here's a step-by-step approach:
  • Identify any factors that can be simplified. For example, in \( \sqrt{12} \), since 12 can be factored into \( 4 \times 3 \), and 4 is a perfect square, it can be broken down into \( \sqrt{4} \cdot \sqrt{3} \).
  • Simplify the perfect square. \( \sqrt{4} = 2 \), thus turning \( \sqrt{12} \) into \( 2\sqrt{3} \).
In our exercise, simplifying \( x \sqrt{12} \) is critical. We break it down as follows:
  • First, factor the 12 under the square root into its simplest form using a perfect square: \( \sqrt{4} \cdot \sqrt{3} \).
  • This simplifies to \( 2 \sqrt{3} \), so our expression becomes \( 2x \sqrt{3} \).
Once you've fully simplified the radicals, the algebraic expression becomes simpler and easier to understand, allowing for further mathematical operations to be performed easily.