Problem 72 - Intermediate Algebra

Question

Use the quotient rule to divide. Then simplify if possible. See Example 5 $$ \frac{\sqrt{270 y^{2}}}{5 \sqrt{3 y^{-4}}} $$

Step-by-Step Solution

Answer

\( \frac{3y^3 \sqrt{10}}{5} \)

Step 1: Apply the Quotient Rule

The given expression is \( \frac{\sqrt{270 y^{2}}}{5 \sqrt{3 y^{-4}}} \). According to the quotient rule, we divide the numerator by the denominator. We'll first separate the constants, coefficients, and variables inside the square roots: \( \sqrt{270 y^2} = \sqrt{270} \sqrt{y^2} \) and \( \sqrt{3 y^{-4}} = \sqrt{3} \sqrt{y^{-4}} \). This gives us \( \frac{\sqrt{270} \sqrt{y^2}}{5 \sqrt{3} \sqrt{y^{-4}}} \).

Step 2: Simplify Square Roots

Simplify each of the square roots: \( \sqrt{270} = \sqrt{3 \cdot 3 \cdot 3 \cdot 10} \), noticing that \( 270 = 3^3 \times 10 \). Therefore, \( \sqrt{270} = 3\sqrt{30} \). Also, \( \sqrt{y^2} = y \), and \( \sqrt{y^{-4}} = y^{-2} \). Thus, the expression inside the square roots simplifies to \( 3y\sqrt{30} \) for the numerator and \( y^{-2}\sqrt{3} \) for the denominator.

Step 3: Simplify the Expression

Now perform the division: \( \frac{3y\sqrt{30}}{5 y^{-2} \sqrt{3}} \). First, divide the coefficients and constants. We have a common factor of \( \sqrt{3} \) in both the numerator and the denominator, reducing to \( \frac{3y\sqrt{10}}{5 y^{-2} \). Next, simplify the variable terms: \( y \div y^{-2} = y^{1 - (-2)} = y^{3} \). So, the expression becomes \( \frac{3y^3 \sqrt{10}}{5} \).

Step 4: Final Simplified Form

Ensuring all simplifications are made, we have the simplified form of the expression: \( \frac{3y^3 \sqrt{10}}{5} \). No further simplifications are possible because \( \sqrt{10} \) contains no perfect square factors to extract, and none of the remaining terms can be reduced further.

Key Concepts

Simplifying Square Roots

Understanding and simplifying square roots is crucial in mathematics, especially in intermediate algebra. A square root essentially asks "what number, when multiplied by itself, gives me the original number?"
To simplify a square root, you aim to find and pull out any perfect square factors. Let's look at an example used in our solution: \( \sqrt{270} \). The first step is to factorize 270 into its prime components:

• 270 = 2 \( \times \) 3 \( \times \) 3 \( \times \) 3 \( \times \) 5.

Notice that there's a pair of 3s, which is a perfect square. Therefore, 3 gets extracted out of the square root, simplifying it to \( 3\sqrt{30} \). Always try to simplify square roots before proceeding with other operations.
The same process can be applied to any number, allowing us to work more effortlessly with these radical expressions.

Variables with Exponents

Variables with exponents are a significant focus in algebra. They represent repeated multiplication. For example, \( y^2 \) means \( y \cdot y \). In our task, the expression includes exponents like \( y^{-4} \). The negative exponent implies that the base, here \( y \), is in the denominator. It's important to use and understand these properties:

• \( y^m \div y^n = y^{m-n} \)
• \( y^{-n} = \frac{1}{y^n} \)

These rules help in the simplification process. When we handled the expression \( y \div y^{-2} \), we applied and simplified it to get \( y^{1 - (-2)} = y^3 \). This step exemplifies how laws of exponents streamline the operation, reducing the chance of errors while making it easier to work with more complex expressions.

Rational Expressions

Rational expressions are ratios of polynomials and can often appear daunting. However, much like fractions in arithmetic, they follow specific rules. Simplifying these expressions often involves reducing both the numerator and the denominator. In our equation, the rational expression was: \( \frac{\sqrt{270 y^2}}{5 \sqrt{3 y^{-4}}} \).
The process of simplification requires applying rules such as:

• Finding common factors
• Simplifying square roots
• Applying exponent rules

By breaking down the expression into easier-to-handle segments, such as separating variables from coefficients and constants, the complexity becomes manageable. Though the expression may seem complex, following systematic rules helps achieve a cleaner, more straightforward result.

Intermediate Algebra

Intermediate algebra bridges basic algebra and advanced mathematics, focusing on developing problem-solving skills and critical thinking. This discipline involves manipulating various mathematical expressions, such as polynomials, radical expressions, and rational expressions.
In exercises like the one given, utilizing techniques like the quotient rule helps divide and simplify expressions. You need to hold a strong grasp over:

• Simplifying radicals
• Working with exponents and their rules
• Simplifying rational expressions

By mastering these concepts, you become well-equipped to tackle complex expressions, making intermediate algebra a crucial stepping stone in higher mathematics. Each problem polished and allows for a deeper understanding of algebra, fostering a foundation for future math challenges.