Problem 51 - College Algebra

Question

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

Step-by-Step Solution

Answer

The simplified expression is \(-w^{\frac{3}{2}} \sqrt{2}\).

Step 1: Simplify Radicals

Begin by simplifying the square roots inside the expression. For \(\sqrt{32}\), it can be rewritten as \(\sqrt{16 \cdot 2} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}\).Similarly, for \(\sqrt{50}\), it can be rewritten as \(\sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}\).

Step 2: Substitute Simplified Radicals Back Into the Expression

Replace the square root terms in the original expression with the simplified forms:\[ w^{\frac{3}{2}} \cdot 4\sqrt{2} - w^{\frac{3}{2}} \cdot 5\sqrt{2} \].

Step 3: Factor Out the Common Term

Notice that \(w^{\frac{3}{2}} \sqrt{2}\) is common in both terms. Factor it out:\[ w^{\frac{3}{2}} \sqrt{2} (4 - 5) \].

Step 4: Simplify the Bracketed Expression

Calculate \(4 - 5\) in the bracketed expression, which simplifies to \(-1\).

Step 5: Final Simplification

Combine the results from the previous steps:\[ w^{\frac{3}{2}} \sqrt{2} \times (-1) = -w^{\frac{3}{2}} \sqrt{2} \].

Key Concepts

Radical Expressions

Radical expressions involve roots, such as square roots or cube roots. In our example, we dealt with square roots, denoted by the symbol \( \sqrt{} \). When simplifying radical expressions, the goal is to find a simpler or more standardized form of the expression.
Breaking down a square root like \( \sqrt{32} \) involves recognizing it as \( \sqrt{16 \cdot 2} \). The number 16 is a perfect square, equal to \( 4^2 \), so \( \sqrt{16} = 4 \). Hence, \( \sqrt{32} \) simplifies to \( 4\sqrt{2} \).
This method applies regardless of the radical involved. The key steps include:

• Identifying perfect square factors inside the radical.
• Extracting these factors, while leaving non-square factors inside the root.
• Ensuring the final expression has no perfect square factors left under the radical.

By simplifying radicals, we make expressions easier to work with, especially in subsequent algebraic operations.

Exponentiation

Exponentiation refers to raising a number or a variable to a certain power. In the given expression, \( w^{\frac{3}{2}} \) is an example of exponentiation. Understanding and working with exponents is crucial for simplifying algebraic expressions.
An exponent like \( \frac{3}{2} \) can be interpreted in two parts:

• The base number, \( w \), raised to the power of 3, which is \( w^3 \).
• The square root of the result, since \( \frac{1}{2} \) indicates a square root.

So, \( w^{\frac{3}{2}} \) is equivalent to \( \sqrt{w^3} \).
For handling fractional exponents:

• The numerator tells you the power.
• The denominator tells you the root.

This allows us to work seamlessly between radical and exponential expressions, streamlining the simplification process within algebra.

Factoring in Algebra

Factoring is breaking down expressions into simpler multiplicative components. It is a powerful algebraic tool that simplifies expressions and solves equations.
In our problem, we noted a common factor: \( w^{\frac{3}{2}} \sqrt{2} \). This was common to both terms in the expression:

• \( w^{\frac{3}{2}} \cdot 4\sqrt{2} \)
• \( w^{\frac{3}{2}} \cdot 5\sqrt{2} \)

Factoring these terms involves pulling out the common factor \( w^{\frac{3}{2}} \sqrt{2} \), which simplifies the expression to \( w^{\frac{3}{2}} \sqrt{2}\cdot(4 - 5) \).
Steps in factoring:

• Identify the greatest common factor (GCF) in all terms.
• Factor out this GCF from each term.
• Express the original expression as a product of this common factor and the reduced expression.

Through factoring, we reduce complexity and make the evaluation and manipulation of algebraic expressions straightforward.