Problem 15

Question

Add or subtract as indicated. Assume that all variables represent positive real numbers. $$ \frac{\sqrt{20 x}}{9}+\sqrt{\frac{5 x}{9}} $$

Step-by-Step Solution

Verified
Answer
The answer is \( \frac{5\sqrt{5x}}{9} \).
1Step 1: Simplify the First Term
The first term is \( \frac{\sqrt{20x}}{9} \). We know that \( \sqrt{20x} = \sqrt{4 \times 5x} = \sqrt{4} \times \sqrt{5x} = 2\sqrt{5x} \). Thus, the first term simplifies to \( \frac{2\sqrt{5x}}{9} \).
2Step 2: Simplify the Second Term
The second term is \( \sqrt{\frac{5x}{9}} \). This can be rewritten by breaking it into two parts: \( \sqrt{\frac{5x}{9}} = \frac{\sqrt{5x}}{\sqrt{9}} = \frac{\sqrt{5x}}{3} \) because \( \sqrt{9} = 3 \).
3Step 3: Find a Common Denominator
The terms are \( \frac{2\sqrt{5x}}{9} \) and \( \frac{\sqrt{5x}}{3} \). The common denominator for 9 and 3 is 9. Convert \( \frac{\sqrt{5x}}{3} \) to a fraction with a denominator of 9 by multiplying the numerator and denominator by 3, resulting in \( \frac{3\sqrt{5x}}{9} \).
4Step 4: Add the Fractions
Now that both terms have the same denominator, combine them: \( \frac{2\sqrt{5x}}{9} + \frac{3\sqrt{5x}}{9} = \frac{(2\sqrt{5x} + 3\sqrt{5x})}{9} = \frac{5\sqrt{5x}}{9} \).

Key Concepts

Simplifying RadicalsCommon DenominatorAddition and Subtraction of Fractions
Simplifying Radicals
When working with radicals, simplification is an important step that helps make expressions easier to work with. A radical expression involves a root symbol, such as a square root (\(\sqrt{}\)). Many radical expressions can be simplified by identifying perfect squares or cubes contained within them. For example, \(\sqrt{20x}\) can be broken down to \(\sqrt{4 \times 5x}\). Since \(\sqrt{4}\) is a perfect square (which equals 2), \(\sqrt{4} \times \sqrt{5x}\) simplifies to \(2\sqrt{5x}\).
This method of simplification helps reduce the complexity of radical expressions and makes it easier to combine them or perform other operations. Remember that simplifying a radical often involves:
  • Breaking down numbers into their prime factors.
  • Identifying and extracting any perfect square factors outside the radical.
Understanding how to effectively simplify radicals is crucial when preparing to perform arithmetic operations on them, such as addition or subtraction.
Common Denominator
Finding a common denominator is essential when adding or subtracting fractions. The process involves converting each fraction so they share the same bottom number, or denominator. This step allows you to directly combine the numerators.
In the example, we began with fractions \(\frac{2\sqrt{5x}}{9}\) and \(\frac{\sqrt{5x}}{3}\). To perform addition, both fractions need the same denominator. The easiest way to find a common denominator is to look for the least common multiple (LCM) of the existing denominators. Here, the LCM of 3 and 9 is 9.
Next, adjust the fractions so that each has this common denominator:
  • Multiply both the numerator and the denominator of \(\frac{\sqrt{5x}}{3}\) by 3, resulting in \(\frac{3\sqrt{5x}}{9}\).
This step is crucial because it sets the stage for adding or subtracting fractions by turning them into compatible terms.
Addition and Subtraction of Fractions
Once fractions have a common denominator, you can easily add or subtract them by combining their numerators while leaving the denominators unchanged. This straightforward process simplifies the addition or subtraction of fraction expressions.
In the given problem, after converting both terms to have a denominator of 9, we have \(\frac{2\sqrt{5x}}{9}\) and \(\frac{3\sqrt{5x}}{9}\). Since they share the same denominator, you can add the numerators directly:
  • The combined expression becomes \(\frac{(2\sqrt{5x} + 3\sqrt{5x})}{9}\).
  • Simplify the numerator by adding the coefficients: \(2\sqrt{5x} + 3\sqrt{5x} = 5\sqrt{5x}\).
Thus, the final result of the operation is \(\frac{5\sqrt{5x}}{9}\).
Being proficient with these operations is vital as it allows you to handle more complex mathematical problems with ease, providing a strong base for advanced studies in algebra and calculus.