Problem 48 - College Algebra

Question

For the following exercises, simplify each expression. $$ q^{\frac{1}{2}} \sqrt{63 p} $$

Step-by-Step Solution

Answer

The simplified expression is \( 3\sqrt{7pq} \).

Step 1: Simplify the Inner Radical

The given expression is \( q^{\frac{1}{2}} \sqrt{63p} \). First, we simplify \( \sqrt{63p} \). Notice that 63 can be factored into 9 and 7, where 9 is a perfect square: \( 63 = 9 \times 7 \). Thus, \( \sqrt{63p} = \sqrt{9 \times 7 \times p} = \sqrt{9} \sqrt{7p} = 3\sqrt{7p} \).

Step 2: Multiply with the Outer Term

Now we substitute the simplified radical back into the expression: \( q^{\frac{1}{2}} 3\sqrt{7p} \). This can be rewritten by multiplying the terms: \( 3 q^{\frac{1}{2}} \sqrt{7p} \).

Step 3: Simplify Powers and Radicals

The expression contains the term \( q^{\frac{1}{2}} \) which is equivalent to \( \sqrt{q} \). Therefore, \( 3 q^{\frac{1}{2}} \sqrt{7p} \) becomes \( 3 \sqrt{q} \sqrt{7p} \). This can be further combined using the property of square roots: \( 3\sqrt{7pq} \).

Key Concepts

Radical Expressions

Radical expressions involve numbers or terms that include a radical sign, commonly the square root. Simplifying radical expressions is an essential skill in algebra and involves rewriting the expression in a simpler form or extracting terms if they are perfect squares. For example, when simplifying \( \sqrt{63p} \), recognizing that 63 is the product of 9 (a perfect square) and 7 allows us to rewrite this expression as \( \sqrt{9 \times 7 \times p} \).
​This can then be simplified further to \( \sqrt{9} \sqrt{7p} = 3 \sqrt{7p} \), where \( 3 \) is extracted outside the radical because \( 9 \) is a perfect square. Understanding and identifying perfect squares is crucial as they help to simplify these types of expressions. It is like peeling back the layers to find a simpler version of the same expression, making calculations and further algebraic manipulation easier.

Exponent Rules

Exponent rules are a key component of simplifying algebraic expressions where powers of numbers or variables are involved. For instance, in the expression \( q^{\frac{1}{2}} \), the exponent \( \frac{1}{2} \) indicates a square root. This is because any term raised to the power of \( \frac{1}{2} \) is essentially the square root of that term. Thus, \( q^{\frac{1}{2}} \) is equivalent to \( \sqrt{q} \).

Using exponent rules:

• Multiplying like bases involves adding their exponents, such as \( a^m \times a^n = a^{m+n} \).
• Dividing like bases involves subtracting the exponents, like \( a^m / a^n = a^{m-n} \).
• And when raising a power to another power, you multiply exponents \((a^m)^n = a^{m \times n}\).

Learning these rules helps to manage expressions that might initially seem complicated by restructuring them into a form that’s easier to compute, as we saw when transforming \( q^{\frac{1}{2}} \) into \( \sqrt{q} \). Understanding how exponents interact with radicals and other factors is fundamental to mastering simplification.

Factoring

Factoring in algebra involves expressing a number or an expression as a product of its divisors, which can often simplify complex expressions. For example, factoring was necessary in our task when we broke down 63 as \( 9 \times 7 \). Recognizing 9 as a perfect square was a key step that allowed us to take \( \sqrt{9} \), which is 3, out of the radical.
Factoring serves multiple purposes:

• Identifies perfect square terms for extraction and simplification.
• Reduces expressions into its simplest terms thus easing further operations.
• Helps in setting up expressions for solving equations, particularly in finding roots.

Approaching expressions with a mindset of finding common factors first can make even the most complex algebraic tasks manageable by reducing them to simpler and recognizable forms. Knowing and applying factoring effectively can greatly ease the simplification of radical expressions, as demonstrated, leading to a more efficient solution.