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Chapter 3: Problem 17

Tell whether each computation is correct or incorrect. $$ \sqrt{2} \cdot \sqrt{3}=\sqrt{6} $$

Short Answer

Expert verified
The computation is correct.

Step by step solution

01

Understand the Problem

The exercise asks to determine if the given computation \( \sqrt{2} \cdot \sqrt{3} = \sqrt{6} \) is correct or not.
02

Review Properties of Radicals

Recall that the product of the square roots of two numbers is equal to the square root of the product of those two numbers: \( \sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b} \).
03

Apply the Property

Using the property from Step 2, compute the product: \( \sqrt{2} \cdot \sqrt{3} = \sqrt{2 \cdot 3} = \sqrt{6} \).
04

Conclusion

Since \( \sqrt{2} \cdot \sqrt{3} = \sqrt{6} \) matches the given computation, the computation is correct.

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Key Concepts

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

Square Roots
Square roots are a fundamental concept in mathematics. The square root of a number is a value that, when multiplied by itself, gives the original number. For instance, the square root of 9 is 3, because \(3 \times 3 = 9\).
Square roots are often represented by the radical symbol \(\sqrt{}\), so \(\sqrt{9} = 3\).
It's important to remember that every positive number has two square roots: one positive and one negative. However, we usually refer to the positive root, known as the principal square root.
Understanding square roots is crucial when dealing with equations or expressions involving radicals.
Multiplication of Radicals
The multiplication of radicals follows a specific property: the product of the square roots of two numbers is equal to the square root of the product of those two numbers.
In other words, \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\).
For example, \(\sqrt{2} \times \sqrt{3} = \sqrt{2 \times 3} = \sqrt{6}\).
This property is incredibly useful for simplifying expressions that involve multiple radicals.
Always ensure to multiply the numbers inside the square roots before taking the square root of the product.
Simplifying Radicals
Simplifying radicals involves reducing the expression under the radical sign to its simplest form.
One common method is to factor the number and pair up identical factors.
For instance, simplifying \(\sqrt{50}\):
  • Factor 50 into its prime factors: \(50 = 2 \times 5 \times 5\)
  • Pair up the identical factors: \(5 \times 5\)
  • Simplify the paired factors: \(\sqrt{50} = \sqrt{2 \times 5 \times 5} = 5\sqrt{2}\)
This makes the expression more manageable. The goal is to make it easy to work with by reducing it to a simpler form.
This technique is valuable in various mathematical problems involving radicals and can make solving equations much smoother.

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Most popular questions from this chapter

Simplify each radical expression. If it is already simplified, say so. $$ \sqrt{17}-\sqrt{30} $$

Without graphing, decide which of these equations represent parallel lines. (Assume that \(q\) is on the horizontal axis.) Explain. \(\begin{array}{ll}{\text { a. } 2 p=3 q+5} & {\text { b. } p=3 q^{2}+5} \\\ {\text { c. } p=1.5 q-7.1} & {\text { d. } p=3 q+3}\end{array}\)

Without computing the value of each pair of numbers, determine which number is greater. For each problem, explain why. $$3^{-1,600} \text { or } 27^{-500}$$

Prove that each number is rational by finding a pair of integers whose ratio, or quotient, is equal to the number. $$ -0.000230 $$

You have worked quite a bit with integer exponents. Exponents can also be fractions. When \(\frac{1}{n}\) is used as an exponent, it means take the nth root. So, for example, $$81^{\frac{1}{2}}=\sqrt{81}=9 \quad(-27)^{\frac{1}{3}}=\sqrt[3]{-27}=-3 \quad 64^{\frac{1}{4}}=\sqrt[4]{64}=4$$ The laws of exponents apply to fractional exponents just as they do to integer exponents. Evaluate each expression without using a calculator. In Parts e-h, use the laws of exponents to help you. \(\begin{array}{llll}{\text { a. } 1.44^{\frac{1}{2}}} & {\text { b. } 125^{\frac{1}{3}}} & {\text { c. }(-32)^{\frac{1}{3}}} & {\text { d. }-32^{\frac{1}{5}}} \\ {\text { e. }\left(3^{\frac{1}{3}}\right)^{3}} & {\text { f. }\left(\frac{9}{25}\right)^{\frac{1}{2}}} & {\text { g. }(64)^{-\frac{1}{3}}} & {\text { h. } 16^{\frac{3}{4}}}\end{array}\)
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