Problem 63 - Algebra for College Students

Question

Sometimes it is more convenient to express a number as a product of a power of 10 and a number that is not between 1 and 10 . For example, suppose that we want to calculate \(\sqrt{640,000}\). We can proceed as follows: $$ \begin{aligned} \sqrt{640,000} &=\sqrt{(64)(10)^{4}} \\ &=\left((64)(10)^{4}\right)^{\frac{1}{2}} \\ &=(64)^{\frac{1}{2}}\left(10^{4}\right)^{\frac{1}{2}} \\ &=(8)(10)^{2} \\ &=8(100)=800 \end{aligned} $$ Compute each of the following without a calculator, and then use a calculator to check your answers. (a) \(\sqrt{49,000,000}\) (b) \(\sqrt{0.0025}\) (c) \(\sqrt{14,400}\) (d) \(\sqrt{0.000121}\) (e) \(\sqrt[3]{27,000}\) (f) \(\sqrt[3]{0.000064}\)

Step-by-Step Solution

Answer

(a) 7000, (b) 0.05, (c) 120, (d) 0.011, (e) 30, (f) 0.04

Step 1: Express the Number for Part (a)

First, express 49,000,000 as a product of a number and a power of 10. The number can be written as \[ 49,000,000 = 49 \times 10^6 \]

Step 2: Apply the Square Root in Part (a)

Apply the square root to the expression from Step 1:\[ \sqrt{49,000,000} = \sqrt{49 \times 10^6} = (\sqrt{49})(\sqrt{10^6}) \] The square root of 49 is 7 and the square root of \(10^6\) is \(10^3\).

Step 3: Multiply the Results for Part (a)

Multiply the results:\[ 7 \times 10^3 = 7 \times 1000 = 7000 \] Thus, \( \sqrt{49,000,000} = 7000 \).

Step 4: Express the Number for Part (b)

Express 0.0025 as a product of a number and a power of 10:\[ 0.0025 = 25 \times 10^{-4} \]

Step 5: Apply the Square Root in Part (b)

Apply the square root:\[ \sqrt{0.0025} = \sqrt{25 \times 10^{-4}} = (\sqrt{25})(\sqrt{10^{-4}}) \] The square root of 25 is 5 and for \(10^{-4}\), it's \(10^{-2}\).

Step 6: Multiply the Results for Part (b)

Multiply the results:\[ 5 \times 10^{-2} = 5 \times 0.01 = 0.05 \] Thus, \( \sqrt{0.0025} = 0.05 \).

Step 7: Express the Number for Part (c)

Express 14,400 as a product of a number and a power of 10:\[ 14,400 = 144 \times 10^2 \]

Step 8: Apply the Square Root in Part (c)

Apply the square root:\[ \sqrt{14,400} = \sqrt{144 \times 10^2} = (\sqrt{144})(\sqrt{10^2}) \] The square root of 144 is 12 and for \(10^2\), it's 10.

Step 9: Multiply the Results for Part (c)

Multiply the results:\[ 12 \times 10 = 120 \] Thus, \( \sqrt{14,400} = 120 \).

Step 10: Express the Number for Part (d)

Express 0.000121 as a product of a number and a power of 10:\[ 0.000121 = 121 \times 10^{-6} \]

Step 11: Apply the Square Root in Part (d)

Apply the square root:\[ \sqrt{0.000121} = \sqrt{121 \times 10^{-6}} = (\sqrt{121})(\sqrt{10^{-6}}) \] The square root of 121 is 11 and for \(10^{-6}\), it's \(10^{-3}\).

Step 12: Multiply the Results for Part (d)

Multiply the results:\[ 11 \times 10^{-3} = 11 \times 0.001 = 0.011 \] Thus, \( \sqrt{0.000121} = 0.011 \).

Step 13: Express the Number for Part (e)

Express 27,000 as a product of a number and a power of 10:\[ 27,000 = 27 \times 10^3 \]

Step 14: Apply the Cube Root in Part (e)

Apply the cube root:\[ \sqrt[3]{27,000} = \sqrt[3]{27 \times 10^3} = (\sqrt[3]{27})(\sqrt[3]{10^3}) \] The cube root of 27 is 3 and for \(10^3\), it's 10.

Step 15: Multiply the Results for Part (e)

Multiply the results:\[ 3 \times 10 = 30 \] Thus, \( \sqrt[3]{27,000} = 30 \).

Step 16: Express the Number for Part (f)

Express 0.000064 as a product of a number and a power of 10:\[ 0.000064 = 64 \times 10^{-6} \]

Step 17: Apply the Cube Root in Part (f)

Apply the cube root:\[ \sqrt[3]{0.000064} = \sqrt[3]{64 \times 10^{-6}} = (\sqrt[3]{64})(\sqrt[3]{10^{-6}}) \] The cube root of 64 is 4 and for \(10^{-6}\), it's \(10^{-2}\).

Step 18: Multiply the Results for Part (f)

Multiply the results:\[ 4 \times 10^{-2} = 4 \times 0.01 = 0.04 \] Thus, \( \sqrt[3]{0.000064} = 0.04 \).

Key Concepts

Square Roots

The square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 49 is 7, because 7 times 7 equals 49. Calculating square roots becomes simpler when dealing with numbers expressed as a product of another number and a power of 10, such as using scientific notation.

• For instance, to find \( \sqrt{49,000,000} \,\), express 49,000,000 as \( 49 imes 10^6 \,\).
• Apply the square root individually: \( (\sqrt{49})(\sqrt{10^6}) = 7 \times 10^3 \,\).

Breaking down the numbers in this manner makes the calculation straightforward, as it isolates simpler numbers, for which square roots are readily known.

Cube Roots

A cube root of a number is a value that, when used three times in a multiplication, results in the original number. The cube root of 27 is 3, since 3 x 3 x 3 = 27. Cube roots can also be simplified by breaking down numbers into a product of integers and powers of 10.

• Consider \( \sqrt[3]{27,000} \,\), where 27,000 can be expressed as \( 27 \times 10^3 \,\).
• Applying the cube root to both components separately gives us \( (\sqrt[3]{27})(\sqrt[3]{10^3}) = 3 \times 10 \,\).

This simplification method helps in recognizing patterns and simplifies calculations involving both large and small numbers.

Scientific Notation

Scientific notation allows large or small numbers to be expressed compactly as a product of a number between 1 and 10 and a power of 10. For example, 0.0025 written in scientific notation is \( 25 \times 10^{-4} \,\). This notation streamlines calculations, especially when dealing with roots.

• To find \( \sqrt{0.0025} \,\):
• First write it as \( 25 \times 10^{-4} \,\) and apply the square root to both parts: \( (\sqrt{25})(\sqrt{10^{-4}}) \,\).

Scientific notation is a vital tool in simplifying these operations into more manageable steps.

Simplifying Expressions

Simplifying expressions involves reducing them to a form that is easier to understand or compute, without changing the value. This process is essential when dealing with roots, as it helps in breaking down complex problems into simpler parts.

• Take 14,400 as an example. Express \( 14,400 \,\) as \( 144 \times 10^2 \,\).
• Apply the square root to each component: \( (\sqrt{144})(\sqrt{10^2}) = 12 \times 10 \,\).
• The simplified result is 120.

By using these simplification techniques, the computations for square and cube roots become more straightforward, aiding clarity and reducing errors in calculations.