Problem 331 - Elementary Algebra

Question

In the following exercises, simplify. $$ \frac{\sqrt{75 r^{3}}}{\sqrt{108 r}} $$

Step-by-Step Solution

Answer

\( \frac{5r}{6} \).

Step 1: Simplify the Radicals

First, break down the radicals into their prime factors. \( \sqrt{75 r^3} = \sqrt{3 \times 5^2 \times r^3} \) and \( \sqrt{108 r} = \sqrt{2^2 \times 3^3 \times r} \).

Step 2: Simplify Each Radicand

Simplify each expression under the square roots. \( \sqrt{75 r^3} = \sqrt{3 \times 5^2 \times r^3} = 5r \sqrt{3r} \) and \( \sqrt{108 r} = \sqrt{2^2 \times 3^3 \times r} = 6 \sqrt{3r} \).

Step 3: Combine the Simplified Expressions

Combine the simplified expressions under a common fraction. \( \frac{5r \sqrt{3r}}{6 \sqrt{3r}} \).

Step 4: Simplify the Fraction

Cancel the common \(\sqrt{3r}\) term in the numerator and the denominator. \( \frac{5r \cancel{\sqrt{3r}}}{6 \cancel{\sqrt{3r}}} = \frac{5r}{6} \).

Key Concepts

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations. In our exercise, expressions include terms like \(75r^3\) and \(108r\). When simplifying, it's important to identify these terms separately before combining them. Each term in an algebraic expression can be simplified, computed, or factored individually.
Understanding how to manipulate these expressions is the foundation of reducing complex equations into simpler forms.

Prime Factorization

Prime factorization involves breaking down a number into its basic building blocks: prime numbers. For instance, to simplify the radicals in the given exercise, we decompose 75 and 108.
\(75 = 3 \times 5^2\)
\(108 = 2^2 \times 3^3\)
Breaking down numbers like this helps us simplify square roots by revealing perfect squares (like \(5^2\) and \(2^2\)).
When a variable raised to a power, like \(r^3\), is included inside the square root, we can factor them similarly: \(r^3 = r^2 \times r\), making the expression easier to handle.

Fraction Simplification

Simplifying fractions means reducing them to their simplest form. Our task simplifies as we cancel common terms in the numerator and denominator.
After simplifying the radicals, our exercise results in the fraction \(\frac{5r\sqrt{3r}}{6\sqrt{3r}}\).
Noticing that \(\sqrt{3r}\) appears in both the numerator and the denominator, we can effectively cancel it out:
\(\frac{5r\cancel{\sqrt{3r}}}{6\cancel{\sqrt{3r}}} = 5r/6\)
This step demonstrates how identifying and canceling common factors can streamline complex fractions to simpler, more manageable forms.

Radicals

Radicals represent root expressions, commonly square roots. In this problem, we deal with \(\sqrt{75r^3}\) and \(\sqrt{108r}\).
Simplifying these requires understanding how to break down and factorize what's under the radical sign. When square roots are simplified, identified perfect squares are taken out of the radical:

• \(\sqrt{75r^3} = 5r\sqrt{3r}\)
• \(\sqrt{108r} = 6\sqrt{3r}\)

The principle here is that taking the square root of a perfect square returns its base. Learning this helps in consistently simplifying radical expressions and makes them easier to work with, especially in more complex algebraic problems.