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Chapter 1: Problem 23

Simplify the given expressions. In each of \(5-9\) and \(12-21,\) the result is an integer. $$\sqrt{1200}$$

Short Answer

Expert verified
The integer result of \(\sqrt{1200}\) is approximately 35.

Step by step solution

01

Express the number as a product of factors

Begin by breaking down the number inside the square root, 1200, into its prime factors. We can start with the smallest prime numbers.- 1200 can be divided by 2 (it is even): \(1200 \div 2 = 600\)- Continue with 600, which can also be divided by 2: \(600 \div 2 = 300\)- Continue dividing by 2: \(300 \div 2 = 150\)- Next, 150 again by 2: \(150 \div 2 = 75\)- Now 75, divide by 3 (next smallest prime): \(75 \div 3 = 25\)- Finally, 25 can be divided by 5 twice: \(25 \div 5 = 5, \quad 5 \div 5 = 1\)Thus, the prime factorization of 1200 is \(2^4 \times 3 \times 5^2\).
02

Simplify using the square root property

Apply the property of square root which states \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\). Use the prime factorization form identified in Step 1.So, \(\sqrt{1200} = \sqrt{2^4 \times 3 \times 5^2} \)Split the square root:- \(\sqrt{1200} = \sqrt{2^4} \times \sqrt{3} \times \sqrt{5^2}\)Calculate the square roots of the perfect powers:- \(\sqrt{2^4} = \sqrt{(2^2)^2} = 2^2 = 4\)- \(\sqrt{3} = \sqrt{3}\) (remains under the root since it's not a perfect square)- \(\sqrt{5^2} = 5\)
03

Combine the factors

Now multiply the simplified factors together:\(4 \times 5 = 20\)Thus, \(\sqrt{1200} = 20 \times \sqrt{3}\).
04

Round to the nearest integer

Since the result requires the expression to evaluate to an integer, we have:- \(\sqrt{3} \approx 1.732\)- Multiply this approximate value by 20: - \(20 \times 1.732 \approx 34.64\)Round 34.64 to the nearest integer, which is 35.

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Key Concepts

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

Prime Factorization
To simplify radicals, understanding prime factorization is crucial. Prime factorization involves expressing a number as a product of its smallest prime numbers. For instance, when simplifying \( \sqrt{1200} \), we start by breaking down 1200 into its prime factors.
  • 1200 is divisible by 2, the smallest prime: \( 1200 \div 2 = 600 \)
  • Continue dividing by 2: \( 600 \div 2 = 300 \)
  • Repeat for 150 and 75: \( 150 \div 2 = 75 \)
  • Switch to the next prime, 3: \( 75 \div 3 = 25 \)
  • Finally, use 5: \( 25 \div 5 = 5 \), and 5 \( \div 5 = 1 \)
Navigating these steps, we determine that the prime factorization of 1200 is \( 2^4 \times 3 \times 5^2 \). Each number is broken down to its smallest building blocks, which helps simplify expressions within a square root.
Square Root Property
The square root property is an essential tool for simplifying radicals. It allows us to separate a product under a square root into the square root of each factor separately. This reduces the complexity of calculations.
The property is mathematical and straightforward:
  • \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \)
Using this property, when simplifying \( \sqrt{1200} \), we utilize the prime factorization result:
  • \( \sqrt{1200} = \sqrt{2^4 \times 3 \times 5^2} \)
  • Separate each factor: \( \sqrt{1200} = \sqrt{2^4} \times \sqrt{3} \times \sqrt{5^2} \)
This method makes it easier to calculate each individual square root, especially when dealing with perfect squares, leading to simplified results.
Perfect Squares
A perfect square is an integer that results from squaring a whole number. Recognizing perfect squares is crucial in simplifying radicals efficiently.
Consider the equation \( \sqrt{2^4 \times 3 \times 5^2} \):
  • \( 2^4 = (2^2)^2 \) is a perfect square, equal to 4 once simplified to \( 2^2 \).
  • \( 5^2 \) simplifies directly to 5.
  • \( 3 \) remains unchanged under the root as it is not a perfect square.
Thus, simplifying \( \sqrt{1200} \) involves dealing directly with perfect squares: \( 4 \times 5 = 20 \), ultimately leading to \( 20 \times \sqrt{3} \). Dealing with perfect squares lets us quickly identify factors that can be removed from under the square root, streamlining the process and making calculations much simpler.

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