Problem 24 - Intermediate Algebra

Question

Simplify each radical expression. All variables represent positive real numbers. $$ \sqrt{75 b^{8} c} $$

Step-by-Step Solution

Answer

The simplified expression is \( 5b^4 \sqrt{3c} \).

Step 1: Factor the Radicand

Identify and factor the expression inside the square root. For the expression \( \sqrt{75 b^8 c} \), factor 75 into prime factors and relate the variables: \( 75 = 3 \times 5^2 \), so the expression becomes \( \sqrt{3 \times 5^2 \times b^8 \times c} \).

Step 2: Apply the Product Rule for Radicals

Use the product rule for radicals which states \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \). Apply this rule to separate the expression into distinct square roots: \( \sqrt{3} \times \sqrt{5^2} \times \sqrt{b^8} \times \sqrt{c} \).

Step 3: Simplify Each Radical

Simplify each individual square root: \( \sqrt{5^2} = 5 \), \( \sqrt{b^8} = b^{4} \) (since \( b^8 = (b^4)^2 \)). The expression is now \( \sqrt{3} \times 5 \times b^4 \times \sqrt{c} \).

Step 4: Combine Simplified Terms

Combine the simplified components into one expression, factoring constants and variable powers outside the square root: \( 5b^4 \sqrt{3c} \).

Key Concepts

Product Rule for Radicals

The product rule for radicals is a handy tool when simplifying radical expressions. This rule states that the square root of a product is the product of the square roots. To put it simply, \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \). This allows us to break down complex expressions into more manageable pieces.

For instance, in the solution of \( \sqrt{75 b^8 c} \), after prime factorizing, we use the product rule to separate this into different square roots: \( \sqrt{3 \times 5^2 \times b^8 \times c} \) becomes \( \sqrt{3} \cdot \sqrt{5^2} \cdot \sqrt{b^8} \cdot \sqrt{c} \).

By applying this rule, each part can be simplified separately, making it easier to achieve the simplest form.

Prime Factorization

Prime factorization involves breaking down a number into its basic building blocks - the prime numbers that multiply together to give the original number. This is foundational for simplification tasks, especially with radicals.

With the number 75, we factor it into \( 3 \times 5^2 \). This means 75 can be expressed as these prime factors which are easier to handle in calculations.

Using prime factorization helps identify perfect squares within a radical. In our exercise, spotting \( 5^2 \) allowed us to easily evaluate \( \sqrt{5^2} = 5 \), greatly simplifying our expression.

Always remember, when dealing with numbers in radical expressions, having them in their prime factorized form can unveil clear simplification paths.

Radicand

The term 'radicand' refers to the number or expression inside a square root symbol, \( \sqrt{} \). Understanding the radicand allows you to simplify square roots effectively. In our exercise, the radicand is \( 75 b^8 c \).

By analyzing and manipulating the radicand, we can apply simplification techniques. First, determine any perfect squares within this radicand to ease the complexity. That's why \( 75 \) was refactored into \( 3 \times 5^2 \). For the variable part, \( b^8 \) is equivalent to \( (b^4)^2 \), showing it can be simplified into \( b^4 \).

Remember, a thoroughly examined radicand gives you the entry point for radical simplification.

Exponent Rules

Exponent rules simplify expressions with powers, especially when they show up within radicals. These rules allow manipulation of powers to simplify roots, such as square roots.

In our given expression, handling \( b^8 \) under a square root comes from understanding that \( b^8 = (b^4)^2 \). Therefore, \( \sqrt{b^8} = b^4 \), because the square root cancels the square power.

Here's a quick reminder of exponent rules that are useful in this context:

• The rule \( (a^m)^n = a^{mn} \) helps in breaking down or combining powers.
• The square root of \( a^2 \) is \( a \), as square roots and squares cancel each other out.

Mastering these exponent rules is crucial for efficiently simplifying expressions with radicals.