Problem 55 - College Algebra

Question

For the following exercises, simplify each expression. $$ w^{\frac{3}{2}} \sqrt{32}-w^{\frac{1}{2}} \sqrt{50} $$

Step-by-Step Solution

Answer

The simplified expression is \( w^{\frac{1}{2}} \sqrt{2} (4w - 5) \).

Step 1: Simplify Square Roots

First, simplify the square roots \( \sqrt{32} \) and \( \sqrt{50} \). Note that \( \sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2} \) and \( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \).

Step 2: Substitute and Simplify When Possible

Substitute the values back into the original expression: \[ w^{\frac{3}{2}} \cdot 4\sqrt{2} - w^{\frac{1}{2}} \cdot 5\sqrt{2}. \] This expression can be factored by taking out the common terms.

Step 3: Factor Out Common Terms

Notice that \( \sqrt{2} \) is a common factor in both terms. Also, \( w^{\frac{1}{2}} \) can be factored out. Therefore, the expression becomes: \[ w^{\frac{1}{2}} \sqrt{2} (4w - 5). \]

Step 4: Final Simplified Expression

After factoring out the common terms, the simplified expression is: \[ w^{\frac{1}{2}} \sqrt{2} (4w - 5). \]

Key Concepts

Factoring

Factoring in algebra involves breaking down expressions into a product of simpler expressions. It's like finding the pieces that multiply together to give the original expression. Factoring can make complex equations more manageable and easier to solve. In the given expression \( w^{\frac{3}{2}} \sqrt{32} - w^{\frac{1}{2}} \sqrt{50} \), we applied factoring by recognizing common terms.
This means:

• We looked for terms that appear in each part of the expression.
• In this example, \( \sqrt{2} \) is a common factor in both terms after simplifying the square roots.
• We also noticed that \( w^{\frac{1}{2}} \) is present in both terms but in different powers.

By factoring out these common elements, we rewrite the expression in a simplified product \( w^{\frac{1}{2}} \sqrt{2} (4w - 5) \). This version is much cleaner and easier to work with. Factoring is powerful because it reveals the structure within an expression and simplifies problem-solving.

Square Roots

Square roots provide a way to express a number that produces a given value when multiplied by itself. The process of simplifying square roots involves finding the largest square factor of the number under the root. This simplifies expressions by allowing terms to be factored and extracted as whole numbers.
Here's how we worked with square roots in the expression:

• For \( \sqrt{32} \), we identified that it can be broken down as \( \sqrt{16 \times 2} \) which equals \( \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2} \) because \( 16 \) is a perfect square.
• Similarly, \( \sqrt{50} \) can be rewritten as \( \sqrt{25 \times 2} \) resulting in \( \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2} \).

This breakdown allows us to simplify complex expressions into more manageable pieces, making it easier to identify common factors. Being comfortable with manipulating square roots is crucial, especially in dealing with properties of radicals and their applications in algebra.

Exponents

Exponents represent repeated multiplication of a number. They are a compact way to show a number being multiplied by itself several times. The expression \( w^{\frac{3}{2}} \) indicates that \( w \) is both squared and taken the square root, due to its fractional form.
Here's a clear look at exponents in the expression:

• \( w^{\frac{3}{2}} \) can be interpreted as \( (w)^{1.5} \) which is equivalent to \( w^{1} \cdot w^{\frac{1}{2}} \) or \( w \cdot \sqrt{w} \).
• \( w^{\frac{1}{2}} \) simply signifies the square root of \( w \).

In algebraic expressions, understanding how these components interact allows us to factor and simplify more precisely. Working with fractional exponents equips us with the ability to handle combinations of powers and roots efficiently, proving essential in simplifying expressions like the one in the problem.