Problem 63 - Precalculus: Functions and Graphs

Question

Simplify the expression, and rationalize the denominator when appropriate. \(\sqrt{9 x^{-4} y^{6}}\)

Step-by-Step Solution

Answer

Simplified expression: \(3x^{-2}y^{3}\). No rationalization needed.

Step 1: Break Down the Square Root

The expression inside the square root is \(9x^{-4}y^{6}\). We can distribute the square root over the product: \(\sqrt{9} \times \sqrt{x^{-4}} \times \sqrt{y^{6}}\).

Step 2: Simplify Each Component

Calculate the square root of each component:- \(\sqrt{9} = 3\)- \(\sqrt{x^{-4}} = x^{-2}\) since \(\sqrt{x^{-4}} = (x^{-4})^{1/2} = x^{-2}\)- \(\sqrt{y^{6}} = y^{3}\) because \(y^{6}\) can be broken into \((y^{3})^{2}\).

Step 3: Combine the Simplified Parts

Combine the results from Step 2 to form the expression: \(3x^{-2}y^{3}\).

Step 4: Rationalize the Expression if Needed

In this case, since there's no square root in the denominator, no additional rationalization is needed.

Key Concepts

Rationalizing the Denominator

Rationalizing the denominator is a technique used to eliminate radicals, such as square roots, from the bottom part (denominator) of a fraction. This makes expressions easier to work with and simplifies them for further mathematical operations. When a rational number is in the denominator, it is preferable in many mathematical contexts to have a whole number or a simpler rational form. Here's how you can achieve that:

• Multiply both the numerator and the denominator by the conjugate of the denominator if it is a binomial involving a square root.
• If it's a single term like \(\frac{1}{\sqrt{a}}\), multiply the numerator and denominator by \(\sqrt{a}\) to get \(\frac{\sqrt{a}}{a}\).

In practice, rationalizing the denominator ensures solutions can be easily compared and further simplified without the complications that radicals introduce in the denominator. It is often applied after simplifying other parts of the expression, much like the exercise did with the expression \(\sqrt{9x^{-4}y^{6}}\). In our specific case, rationalization wasn't needed as no radicals were present in the denominator.

Square Roots

The concept of square roots is pivotal for simplifying expressions involving powers. A square root of a number is a value that, when multiplied by itself, gives that number. For example, the square root of 9 is 3 because \(3 \times 3 = 9.\)Understanding how to work with square roots can help make expressions more manageable. Here's a breakdown of handling square roots in the expression \(\sqrt{9 x^{-4} y^{6}}\):

• Distributing the square root over multiplication: You can split the square root of a product into the product of square roots, making it easier to tackle complex expressions. For instance, \(\sqrt{9 x^{-4} y^{6}}\) becomes \(\sqrt{9} \times \sqrt{x^{-4}} \times \sqrt{y^{6}}\).
• Evaluating each square root: Simplify the expression by calculating each square root. \(\sqrt{9}=3\), \(\sqrt{x^{-4}} = x^{-2}\) using the property \((x^{m})^n = x^{m \cdot n}\), and \(\sqrt{y^{6}} = y^{3}\).

This basic understanding helps in simplifying complex expressions efficiently and correctly by breaking them down into more manageable parts.

Exponent Rules

Exponent rules are essential for working with expressions that involve powers and roots, such as the one in the exercise, \(\sqrt{9x^{-4}y^{6}}\). Here are the core exponent rules that are helpful:

• Product of Powers Rule: When multiplying like bases, add the exponents. For example, \(a^m \times a^n = a^{m+n}.\)
• Power of a Power Rule: When you raise a power to another power, multiply the exponents. For example, \((a^m)^n = a^{m \cdot n}.\)
• Negative Exponent Rule: A negative exponent indicates the reciprocal of the base raised to the positive exponent: \(a^{-m} = \frac{1}{a^m}.\)

In the given exercise, these rules are applied when simplifying components inside the square root. For instance, \(\sqrt{x^{-4}}\) is simplified to \(x^{-2}\) by expressing it as \((x^{-4})^{1/2}= x^{-2}\), applying the power of a power rule. Understanding these rules makes it easier to approach and simplify complex expressions with confidence, enabling students to tackle potential challenges in higher mathematics.