Problem 33

Question

Rationalize the denominator and write each fraction in simplest form. All variables represent positive numbers. \(\frac{4}{\sqrt{z}+8}\)

Step-by-Step Solution

Verified
Answer
The rationalized and simplest form is \( \frac{4\sqrt{z} - 32}{z - 64} \).
1Step 1: Understand the Problem
We are given the expression \( \frac{4}{\sqrt{z}+8} \) and need to rationalize the denominator, which means we must eliminate the square root from the denominator.
2Step 2: Identify the Conjugate
The conjugate of the denominator \( \sqrt{z} + 8 \) is \( \sqrt{z} - 8 \). We will use this to rationalize the denominator.
3Step 3: Multiply by the Conjugate
Multiply both the numerator and the denominator by the conjugate of the denominator: \( \frac{4}{\sqrt{z}+8} \times \frac{\sqrt{z}-8}{\sqrt{z}-8} \).
4Step 4: Distribute in the Numerator
Distribute 4 in the numerator: \( 4(\sqrt{z}-8) = 4\sqrt{z} - 32 \).
5Step 5: Apply Difference of Squares in the Denominator
Apply the formula \((a+b)(a-b) = a^2 - b^2\) to the denominator: \( (\sqrt{z}+8)(\sqrt{z}-8) = z - 64 \). Thus, the denominator becomes \( z - 64 \).
6Step 6: Write the Rationalized Expression
The fraction becomes \( \frac{4\sqrt{z} - 32}{z - 64} \). This is the expression with the rationalized denominator.
7Step 7: Simplify If Possible
Check if the fraction can be simplified further. In this case, there is no common factor to simplify \( \frac{4\sqrt{z} - 32}{z - 64} \) further.

Key Concepts

Simplest FormDifference of SquaresConjugate in Algebra
Simplest Form
When working with fractions, transforming them into their simplest form is a way to make them easier to handle and more understandable. This process involves reducing the fraction so that both the numerator and the denominator are as small as possible while still retaining the same value as the original fraction.
  • Firstly, check for any common factors in the numerator and the denominator. If both parts of the fraction share a factor, you can divide them by this number.
  • Next, ensure that the fraction cannot be reduced any further by checking again for common factors.
In the case of the original exercise, where we ended up with \( \frac{4\sqrt{z} - 32}{z - 64} \), we must check if these terms can be further simplified. Since there are no common factors, this is already in its simplest form. Understanding how to simplify expressions is crucial because it aids in comparison, arithmetic operations, and makes the expression more readable.
Difference of Squares
The Difference of Squares is a mathematical technique used to simplify expressions. When applied correctly, it results in making expressions more manageable, especially when dealing with polynomials or algebraic fractions.
  • The formula for the Difference of Squares is \((a+b)(a-b) = a^2 - b^2\). This formula shows how the sum and difference of the same two terms can give you a simplified square result.
  • In quadratic equations, recognizing opportunities to apply this formula can drastically simplify the solving process.
In our exercise, \((\sqrt{z}+8)(\sqrt{z}-8)\) on the denominator became \(z - 64\), illustrating this formula. The process involves squaring the terms \(\sqrt{z}\) (which gives \(z\)), and \(8\) (which gives \(64\)), then subtracting. Using this method simplifies the algebraic expression, crucial for rationalizing denominators.
Conjugate in Algebra
Understanding how to use a conjugate is essential in algebra, particularly when you need to rationalize denominators. A conjugate refers to a binomial formed by changing the sign between two terms.
  • For instance, the conjugate of \(\sqrt{z}+8\) is \(\sqrt{z}-8\).
  • Multiplying by a conjugate is often used to eliminate radicals in the denominator.
Applying the concept of a conjugate in our exercise was pivotal. By multiplying both the numerator and denominator by \(\sqrt{z}-8\), we transformed \(\frac{4}{\sqrt{z}+8}\) into \(\frac{4(\sqrt{z}-8)}{(\sqrt{z})^2-8^2}\), simplifying it significantly. This step is crucial because it helps to structure problems in a form that is easier to resolve, allowing further algebraic operations without the complication of radical terms in denominators.