Problem 178
Question
In the following exercises, simplify. $$ \sqrt{48}+2 \sqrt{3} $$
Step-by-Step Solution
Verified Answer
\[\root{48} + 2 \root{3} = 6 \root{3}\]
1Step 1 - Simplify the Square Root
First, simplify \(\root{48}\). Find the prime factorization of 48 to identify perfect squares. \(48 = 2 \times 24 = 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3\). Therefore, \(\root{48} = \root{2^4 \times 3} = \root{2^4} \times \root{3} = 2^2 \times \root{3} = 4 \root{3}\).
2Step 2 - Combine Like Terms
Now combine like terms: \[ \root{48} + 2 \root{3} = 4 \root{3} + 2 \root{3}. \] Since both are like terms, simply add the coefficients: \[ 4 \root{3} + 2 \root{3} = (4 + 2) \root{3} = 6 \root{3}. \]
Key Concepts
Prime FactorizationPerfect SquaresCombining Like Terms
Prime Factorization
Prime factorization is the process of breaking down a composite number into its prime factors. A prime number is one that only has two distinct positive divisors: 1 and itself. To simplify square roots, it's essential to find the prime factors of the number under the square root.
- For example, to factorize 48, we start by dividing by the smallest prime number, 2:
- 48 ÷ 2 = 24
- 24 ÷ 2 = 12
- 12 ÷ 2 = 6
- 6 ÷ 2 = 3
- 48 = 2 × 2 × 2 × 2 × 3 = 2^4 × 3
Perfect Squares
Perfect squares are numbers that can be expressed as the product of an integer with itself. For example, 1, 4, 9, 16, 25, and 36 are perfect squares since:
- 1 = 1 × 1
- 4 = 2 × 2
- 9 = 3 × 3
- 16 = 4 × 4
- 25 = 5 × 5
- 36 = 6 × 6
- For instance, \( \root{48} = \root{2^4 \times 3} \). Since \( 2^4 \) is a perfect square (\( 2^4 = (2^2)^2 \)), we can simplify it to 4.
- Therefore, \( \root{48} = \root{2^4 \times 3} = 4\root{3} \).
Combining Like Terms
Combining like terms is an essential step in simplifying algebraic expressions. In the context of square roots, like terms are those that have the same radical part. For example:
- In the expression \( \root{48} + 2 \root{3} \), we simplify \( \root{48} \) to \( 4 \root{3} \), giving us \( 4 \root{3} + 2 \root{3} \).
- Since the terms \( 4 \root{3} \) and \( 2 \root{3} \) have the same radical part (\( \root{3} \)), they are like terms and can be combined.
- Add the coefficients (numbers in front of the radicals): \( 4 + 2 = 6 \).
- So, \( 4 \root{3} + 2 \root{3} = 6 \root{3} \).
Other exercises in this chapter
Problem 176
In the following exercises, simplify. $$ 11 \sqrt{2 r s}-9 \sqrt{2 r s}+3 \sqrt{2 r s} $$
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In the following exercises, simplify. $$ \sqrt{80}-3 \sqrt{5} $$
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In the following exercises, simplify. $$ \sqrt{28}-4 \sqrt{7} $$
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