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Chapter 7: Problem 14

Simplify each expression. $$ \sqrt{28} $$

Short Answer

Expert verified
The simplified expression for \( \sqrt{28} \) is \( 2\sqrt{7} \).

Step by step solution

01

Factor the Number Under the Square Root

First, identify the factors of 28. The number 28 can be factored into prime numbers as: \(28 = 2 \times 2 \times 7\).
02

Pair the Prime Factors

Next, look for pairs of the same factor. From the factors of 28, \(2 \times 2\) can form a pair.
03

Simplify the Square Root

For each pair of identical factors inside the square root, you can bring one factor outside. Hence, \( \sqrt{28} = \sqrt{2 \times 2 \times 7} = 2\sqrt{7} \) by taking \( \, 2 \,\) out of the square root.

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Key Concepts

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

Prime Factorization
Prime factorization is the process of expressing a number as a product of its prime numbers. Prime numbers are integers greater than 1 that have only two divisors: 1 and themselves. For example, the number 28 can be broken down into smaller parts by dividing it by prime numbers until only prime numbers are left. This is how we do it:
  • Divide 28 by 2, the smallest prime number: 28 梅 2 = 14. So, 28 is 2 times 14.
  • Next, divide 14 by 2: 14 梅 2 = 7. Now, 14 is 2 times 7.
  • Finally, 7 is already a prime number, so the factorization stops here.
Now we see that 28 is written as: \(28 = 2 \times 2 \times 7 \). Remember, using prime factorization makes it easier to understand the number's structure, and it is a crucial step when simplifying square roots to find pairs of identical factors.
Algebraic Expressions
Algebraic expressions are combinations of letters and numbers using algebraic operations (addition, subtraction, multiplication, division, and exponentiation). When simplifying square roots, we often use algebra to express the square root in its simplest form by factoring it algebraically. Consider the expression \( \sqrt{2 \times 2 \times 7} \). This can be rewritten using algebra as:
  • Identify pairs of factors: We have two 2's that can make a pair.
  • For every pair of identical factors inside the square root, one can be taken outside: This allows \( 2 \) to come out of the square root, turning \( \sqrt{2 \times 2} \) into \( 2 \).
  • Thus, the expression becomes \( 2\sqrt{7} \), which is its simplest form.
Simplifying square roots is a practical example of using algebraic expressions. It helps you find an expression easier to understand or further use in calculations.
Mathematics Education
Mathematics education aims to develop a deep understanding of mathematical concepts and enhance problem-solving skills. Simplifying expressions like square roots is not just about getting the right answer. It helps in building:
  • Critical Thinking: Students learn to analyze and break down numbers, identifying patterns and solutions.
  • Conceptual Understanding: By simplifying square roots, students see how factors work together, which is vital in other math areas like algebra and geometry.
  • Practical Skills: It provides hands-on experience with concepts that are essential for advanced mathematics, such as calculus or statistics.
Through these exercises, learners gain confidence in handling numbers and algebraic expressions with ease, providing a foundation for educational achievement and everyday problem-solving.

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Most popular questions from this chapter

Simplify each expression. Write the answers without negative exponents. All variables represent positive real numbers. See Example 8. $$ \frac{2^{5 / 6} 2^{1 / 3}}{2^{1 / 2}} $$

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt[3]{x+4}=-1 $$

The frequency of vibration of a string varies directly as the square root of the tension and inversely as the length of the string. Suppose a string 2.5 feet long, under a tension of 16 pounds, vibrates 25 times per second. Find \(k\) the constant of proportionality.

Multiply. $$ \sqrt{-2} \sqrt{-12} $$

Find the missing lengths in each triangle. Give the exact answer and then an approximation to two decimal places, when appropriate. See Example \(6 .\) In a \(30^{\circ}-60^{\circ}-90^{\circ}\) right triangle, the length of the hypotenuse is \(12 \sqrt{3}\) inches. Find the length of the leg opposite the \(30^{\circ}\) angle and the length of the leg opposite the \(60^{\circ}\) angle. Give the exact answer and then an approximation to two decimal places, when appropriate.
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