Problem 198
Question
In the following exercises, simplify. $$ \sqrt{48 b^{5}}-\sqrt{75 b^{5}} $$
Step-by-Step Solution
Verified Answer
\( - \sqrt{3} \cdot b^{5/2} \)
1Step 1 - Simplify Inside the Radicals
Write the expressions inside the square roots using their prime factors. For \( \sqrt{48 b^5} \), we have \( \sqrt{16 \cdot 3 \cdot b^5} \) since 48 = 16 \cdot 3 and for \( \sqrt{75 b^5} \), we have \( \sqrt{25 \cdot 3 \cdot b^5} \) since 75 = 25 \cdot 3.
2Step 2 - Separate the Radicals
Use the property \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \) to separate the radicals. Thus, \( \sqrt{48 b^5} = \sqrt{16} \cdot \sqrt{3} \cdot \sqrt{b^5} = 4 \sqrt{3} \cdot b^{5/2} \). Similarly, \( \sqrt{75 b^5} = \sqrt{25} \cdot \sqrt{3} \cdot \sqrt{b^5} = 5 \sqrt{3} \cdot b^{5/2} \).
3Step 3 - Substitute and Combine Like Terms
Now that the radicals are simplified, substitute them back into the original expression: \( 4 \sqrt{3} \cdot b^{5/2} - 5 \sqrt{3} \cdot b^{5/2} \). Factor out the common term \( \sqrt{3} \cdot b^{5/2} \): \( \sqrt{3} \cdot b^{5/2} (4 - 5) = - \sqrt{3} \cdot b^{5/2} \).
Key Concepts
prime factorizationproperties of square rootscombining like terms
prime factorization
Prime factorization is the process of breaking down a number into its prime factors, which are the prime numbers that multiply together to give the original number. It’s crucial for simplifying expressions inside square roots.
For example, in the expression \( \sqrt{48 b^5} - \sqrt{75 b^5} \), we start by breaking down 48 and 75 into their prime factors:
48 can be factored as 16 \cdot 3, where 16 = 2^4 and 3 is already a prime number.
So, we write 48 as \(2^4 \cdot 3\).
Similarly, 75 can be factored as 25 \cdot 3, where 25 = 5^2 and 3 is already a prime number.
So, we write 75 as \(5^2 \cdot 3\).
This makes the radicals easier to simplify since we can clearly see the pairs of numbers we can take outside the square root.
Prime factorization streamlines the process of simplifying radicals and is especially useful for managing complex expressions with variables like \( b^5 \).
For example, in the expression \( \sqrt{48 b^5} - \sqrt{75 b^5} \), we start by breaking down 48 and 75 into their prime factors:
48 can be factored as 16 \cdot 3, where 16 = 2^4 and 3 is already a prime number.
So, we write 48 as \(2^4 \cdot 3\).
Similarly, 75 can be factored as 25 \cdot 3, where 25 = 5^2 and 3 is already a prime number.
So, we write 75 as \(5^2 \cdot 3\).
This makes the radicals easier to simplify since we can clearly see the pairs of numbers we can take outside the square root.
Prime factorization streamlines the process of simplifying radicals and is especially useful for managing complex expressions with variables like \( b^5 \).
properties of square roots
Understanding the properties of square roots helps in simplifying expressions. One key property is: \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \). This property is used to separate the terms inside a square root for easier manipulation.
In the given exercise, we use this property as follows:
For \( \sqrt{48 b^5} \), we write it as \( \sqrt{16 \cdot 3 \cdot b^5} \), then separate these into individual square roots: \( \sqrt{16} \cdot \sqrt{3} \cdot \sqrt{b^5}\). Since \( \sqrt{16} = 4\) and \( \sqrt{b^5} = b^{5/2} \), we get: \( 4 \sqrt{3} \cdot b^{5/2} \).
For \( \sqrt{75 b^5} \), we do the same: \( \sqrt{25 \cdot 3 \cdot b^5}\) becomes \( \sqrt{25} \cdot \sqrt{3} \cdot \sqrt{b^5}\). Since \( \sqrt{25} = 5\), we get: \( 5 \sqrt{3} \cdot b^{5/2} \).
These properties allow us to rewrite the original problem in a simpler form and ultimately combine like terms.
In the given exercise, we use this property as follows:
For \( \sqrt{48 b^5} \), we write it as \( \sqrt{16 \cdot 3 \cdot b^5} \), then separate these into individual square roots: \( \sqrt{16} \cdot \sqrt{3} \cdot \sqrt{b^5}\). Since \( \sqrt{16} = 4\) and \( \sqrt{b^5} = b^{5/2} \), we get: \( 4 \sqrt{3} \cdot b^{5/2} \).
For \( \sqrt{75 b^5} \), we do the same: \( \sqrt{25 \cdot 3 \cdot b^5}\) becomes \( \sqrt{25} \cdot \sqrt{3} \cdot \sqrt{b^5}\). Since \( \sqrt{25} = 5\), we get: \( 5 \sqrt{3} \cdot b^{5/2} \).
These properties allow us to rewrite the original problem in a simpler form and ultimately combine like terms.
combining like terms
Combining like terms is an essential step in simplifying algebraic expressions. It involves adding or subtracting terms that have the same variables raised to the same power.
In our exercise, after simplifying the radicals, we reach the expression: \( 4 \sqrt{3} \cdot b^{5/2} - 5 \sqrt{3} \cdot b^{5/2} \).
Here, both terms contain the factor \( \sqrt{3} \cdot b^{5/2} \).
We can factor this common term out: \( \sqrt{3} \cdot b^{5/2} (4 - 5) \).
Subtracting 4 from 5 gives us -1, so the expression simplifies to: \( - \sqrt{3} \cdot b^{5/2} \).
This process of combining like terms makes complex expressions easier to handle and results in a cleaner, more simplified answer. Always look for terms that can be combined to simplify your work.
In our exercise, after simplifying the radicals, we reach the expression: \( 4 \sqrt{3} \cdot b^{5/2} - 5 \sqrt{3} \cdot b^{5/2} \).
Here, both terms contain the factor \( \sqrt{3} \cdot b^{5/2} \).
We can factor this common term out: \( \sqrt{3} \cdot b^{5/2} (4 - 5) \).
Subtracting 4 from 5 gives us -1, so the expression simplifies to: \( - \sqrt{3} \cdot b^{5/2} \).
This process of combining like terms makes complex expressions easier to handle and results in a cleaner, more simplified answer. Always look for terms that can be combined to simplify your work.
Other exercises in this chapter
Problem 196
In the following exercises, simplify. $$ \frac{1}{3} \sqrt{24}+\frac{1}{4} \sqrt{54} $$
View solution Problem 197
In the following exercises, simplify. $$ \sqrt{72 a^{5}}-\sqrt{50 a^{5}} $$
View solution Problem 200
In the following exercises, simplify. $$ \sqrt{96 d^{9}}-\sqrt{24 d^{9}} $$
View solution Problem 201
In the following exercises, simplify. $$ 9 \sqrt{80 p^{4}}-6 \sqrt{98 p^{4}} $$
View solution