Problem 267 - Elementary Algebra

Question

In the following exercises, simplify. (a) \(\quad 7(-2-\sqrt{11})\) (b)\(\sqrt{7}(6-\sqrt{14})\)

Step-by-Step Solution

Answer

(a) -14 - 7\sqrt{11}, (b) 6\sqrt{7} - 7\sqrt{2}

Step 1: Distribute (a)

To simplify the expression, distribute the 7 to both terms inside the parentheses. \[ 7(-2 - \sqrt{11}) = 7 \cdot (-2) + 7 \cdot (-\sqrt{11}) \]

Step 2: Simplify (a)

Multiply 7 by -2 and 7 by -\sqrt{11}. \[ 7(-2) + 7(-\sqrt{11}) = -14 - 7\sqrt{11} \] The simplified form of the expression is \(-14 - 7\sqrt{11}\).

Step 1: Distribute (b)

Distribute the \sqrt{7} to both terms inside the parentheses. \[ \sqrt{7}(6 - \sqrt{14}) = \sqrt{7} \cdot 6 - \sqrt{7} \cdot \sqrt{14} \]

Step 2: Simplify Products (b)

Multiply \sqrt{7} by 6 and simplify \sqrt{7} \cdot \sqrt{14}. \[ \sqrt{7} \cdot 6 - \sqrt{7 \cdot 14} = 6\sqrt{7} - \sqrt{98} \]

Step 3: Simplify Radicals (b)

Simplify \sqrt{98}. \[ \sqrt{98} = \sqrt{2 \cdot 49} = 7\sqrt{2} \] Substituting this back in gives: \[ 6\sqrt{7} - 7\sqrt{2} \]

Key Concepts

Distributive Property

The distributive property is one of the most important properties in algebra. It helps you to simplify expressions where you need to distribute a multiplication over addition or subtraction. For instance, in the expression \(a(b + c)\), you can distribute the multiplication as follows: \(a(b + c) = ab + ac\). This means you multiply the outside term by each term inside the parentheses.
In the exercise provided, the distributive property was applied to expressions with both integers and radicals. Here’s a step-by-step look:

• For (a), distribute the 7 to both terms inside the parentheses: \[ 7(-2 - \sqrt{11}) = 7 \cdot (-2) + 7 \cdot (-\sqrt{11})\]
• For (b), distribute the \sqrt{7} to each term inside the parentheses: \[ \sqrt{7}(6 - \sqrt{14}) = \sqrt{7} \cdot 6 - \sqrt{7} \cdot \sqrt{14}\]

The distributive property is especially helpful when simplifying more complex algebraic expressions, such as those involving radicals.

Simplifying Radicals

Simplifying radicals involves breaking down a square root into its simplest form. This can make it easier to work with and further simplify expressions. Here's how you can do it:

• **Understand Prime Factors**: Radicals are simplified by factoring the number under the square root into its prime factors. For example, to simplify \sqrt{98}, you first factorize 98 as \(2 \cdot 49\).
• **Identify Perfect Squares**: Look for perfect square factors within the radical. For \sqrt{98}, notice that 49 is a perfect square since \sqrt{49} = 7.
• **Rewrite the Radical**: Break the radical into a product of the square root of the perfect square and the square root of the remaining factor: \sqrt{98} = \sqrt{49 \cdot 2} = 7 \sqrt{2}.

Applying this process to the exercise:

• For (b) in step 3, we simplify \sqrt{98} as: \[ \sqrt{98} = \sqrt{2 \cdot 49} = 7 \sqrt{2} \]Substituting back gives: \[ 6 \sqrt{7} - 7 \sqrt{2} \]

Grasping this concept will make simplifying other radicals easier!

Multiplying Radicals

Multiplying radicals may seem tricky, but it follows simple steps if you understand the properties of square roots. Whenever you multiply two square roots, you can combine them under a single radical sign:

• **Multiplication Rule**: The rule is \[ \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \]. You simply multiply the numbers inside the radicals and take the square root of the product.
• **Example**: For expression (b) \sqrt{7} \cdot \sqrt{14}. You apply the rule: \[ \sqrt{7} \cdot \sqrt{14} = \sqrt{7 \cdot 14} = \sqrt{98} \].

After this step, you simplify the radicals as shown in previous sections.
In the exercise provided:

• First, distribute \sqrt{7} to each term: \[ \sqrt{7}(6 - \sqrt{14}) = \sqrt{7} \cdot 6 - \sqrt{7} \cdot \sqrt{14} = 6 \sqrt{7} - \sqrt{98} \]
• Next, simplify \sqrt{98} as \[7 \sqrt{2} \] to get the final expression as: \[ 6 \sqrt{7} - 7 \sqrt{2} \]

Understanding how to multiply and simplify radicals is essential for mastering algebra, and with practice, it will become second nature!