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Chapter 9: Problem 129

Simplify. $$ \sqrt{\frac{75 r^{9}}{s^{8}}} $$

Short Answer

Expert verified
\( \frac{5 r^4 \sqrt{3r}}{s^4} \)

Step by step solution

01

- Simplify the Radicand

The expression inside the square root is \ \( \frac{75 r^9}{s^8} \). Start by breaking down 75 into its prime factors: \ \( 75 = 3 \times 5^2 \). So the expression becomes \ \( \sqrt{ \frac{3 \times 5^2 \times r^9}{s^8} } \).
02

- Separate the Expression

Split the expression under the square root into separate square roots for each term: \ \( \sqrt{ \frac{3 \times 5^2 \times r^9}{s^8} } = \sqrt{3} \times \sqrt{5^2} \times \sqrt{r^9} \times \sqrt{ \frac{1}{s^8} } \).
03

- Simplify Each Square Root

Simplify each component: \ \( \sqrt{3} \) remains as \ \( \sqrt{3} \). \ \( \sqrt{5^2} = 5 \). \ \( \sqrt{r^9} \) can be simplified because \ \( r^9 = (r^4)^2 \times r \Rightarrow \sqrt{r^9} = \sqrt{(r^4)^2 \times r} = r^4 \sqrt{r} \). \ \( \sqrt{ \frac{1}{s^8} } = \frac{1}{\sqrt{s^8}} = \frac{1}{s^4} \).
04

- Combine Simplified Components

Combine all the simplified parts: \ \( \sqrt{3} \times 5 \times r^4 \sqrt{r} \times \frac{1}{s^4} = 5 r^4 \sqrt{3r} \times \frac{1}{s^4} \).
05

- Write the Final Expression

Combine into a single fraction: \ \( 5 r^4 \sqrt{3r} / s^4 \). Thus, the simplified form of the original expression is \ \( \frac{5 r^4 \sqrt{3r}}{s^4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simplifying Radicands
When we simplify radicands, we aim to make the expression inside a square root as simple as possible. The radicand is the number or expression inside the radical symbol, like the part inside the square root \( \sqrt{ \frac{75 r^9}{s^8} } \) in our problem. Simplifying it is the first crucial step.

Let's break it down:
  • First, we start by expressing each component in terms of its prime factors. For example, 75 can be written as \( 3 \times 5^2 \).
  • Next, any variables with exponents must be handled carefully. For example, \( r^9 \) means \( r \) raised to the power of 9.
  • Lastly, we factor everything under the square root to make subsequent steps easier.
So, simplifying the radicand means breaking down everything inside the square root to its simplest form.
Prime Factorization
Prime factorization helps in simplifying the radicands by breaking down numbers into their basic prime factors. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. To factorize 75, we find its prime components: \( 75 = 3 \times 5^2 \).

Let鈥檚 see how it works:
  • First, identify the smallest prime number that divides 75. Here, it is 3.
  • Next, we divide 75 by 3, resulting in 25. Break down 25 into its prime factors, which are \( 5^2 \).
  • Thus, \( 75 = 3 \times 5^2 \), and we use this in the simplification process.
Remember, prime factorization is crucial because it simplifies the components under the square root, making subsequent steps easier to handle.
Square Roots
Understanding square roots is foundational in algebra. A square root of a number is a value that, when multiplied by itself, gives the original number. For instance, \( \sqrt{25} = 5 \) because \( 5 \times 5 = 25 \).

Let's delve into our exercise:
  • For \( \sqrt{5^2} = 5 \), it directly reduces because squaring 5 i.e., \(5 \times 5 \), under the radical sign gives back 25.
  • Dealing with variables, such as \( r^9 \), is similar. If \( r^9 \) is written as \( (r^4)^2 \times r \, the square root becomes \ r^4 \sqrt{r} \).
  • An inverse relationship exists with denominators like \( \sqrt{ \frac{1}{s^8} } = \frac 1 { \sqrt{s^8} } = \frac 1 {s^4} \).
Combining these ideas allows us to simplify even complex algebraic square roots.

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