Problem 266 - Elementary Algebra

Question

In the following exercises, simplify. (a) \(5(3-\sqrt{7})\) (b) \(\sqrt{3}(4-\sqrt{15})\)

Step-by-Step Solution

Answer

(a) 15 - 5\sqrt{7}, (b) 4\sqrt{3} - 3\sqrt{5}

Step 1: Simplify Expression (a)

Distribute the 5 inside the parentheses for the expression \(5(3 - \sqrt{7})\).

Step 2: Perform Distribution for (a)

Multiply 5 by each term inside the parentheses separately: \(5 \cdot 3 - 5 \cdot \sqrt{7}\).

Step 3: Simplify Product for (a)

Calculate each multiplication: \(15 - 5\sqrt{7}\). The simplified form of expression (a) is \(15 - 5\sqrt{7}\).

Step 4: Simplify Expression (b)

Distribute the \(\sqrt{3}\) inside the parentheses for the expression \(\sqrt{3}(4 - \sqrt{15})\).

Step 5: Perform Distribution for (b)

Multiply \(\sqrt{3}\) by each term inside the parentheses separately: \(\sqrt{3} \cdot 4 - \sqrt{3} \cdot \sqrt{15}\).

Step 6: Simplify Product for (b)

Calculate each multiplication: \(4\sqrt{3} - \sqrt{3 \cdot 15}\).

Step 7: Simplify Radicals for (b)

Simplify the radical \(\sqrt{3 \cdot 15}\) to \(\sqrt{45}\), which further simplifies to \(3\sqrt{5}\). Your expression for (b) is now \(4\sqrt{3} - 3\sqrt{5}\).

Key Concepts

Distributive Property

The distributive property is a key concept in algebra that allows you to multiply a term by each term within parentheses. This property makes it easier to simplify expressions and solve equations. Let's break down how it works with an example.

In the exercise, we have the expression: \(5(3 - \sqrt{7})\).
This means we need to distribute the 5 to both 3 and \(\sqrt{7}\).
So, we do:\(5 \cdot 3 - 5 \cdot \sqrt{7}\).

After multiplying, we get:\(15 - 5\sqrt{7}\).

This is how we used the distributive property to simplify the expression.

Simplifying Radicals

Simplifying radicals involves breaking down a radical expression into its simplest form. This can sometimes mean factoring out perfect squares or other techniques.

In the exercise expression:\(\sqrt{3}(4 - \sqrt{15})\), after distributing \(\sqrt{3}\), we get:\(4\sqrt{3} - \sqrt{3} \cdot \sqrt{15}\).

Next, we simplify \(\sqrt{3} \cdot \sqrt{15}\).
Using the multiplication property of radicals, we get:\(\sqrt{3 \cdot 15}\) or \(\sqrt{45}\).

Finally, simplify \(\sqrt{45}\).
Since \(45 = 9 \cdot 5\) and \(\sqrt{9} = 3\), we get:\(3\sqrt{5}\).

Putting it all together, the expression is simplified to:\(4\sqrt{3} - 3\sqrt{5}\).

Multiplication of Radicals

Multiplying radicals can seem tricky, but it's simple once you understand the basic rules. When you multiply two radicals, you can combine them under a single square root.

For example, in the expression:\(\sqrt{3} \cdot \sqrt{15}\), we can combine them as:\(\sqrt{3 \cdot 15}\).

This gives us:\(\sqrt{45}\), which can then be simplified further, as mentioned above.

Remember, the key point is to look for factors that are perfect squares inside the radical so you can simplify the expression further. Using these techniques can help make any algebra problem involving radicals much easier to handle.