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\(15-24\) . Evaluate each expression. $$ \begin{array}{llll}{\text { (a) } \sqrt{\frac{4}{9}}} & {\text { (b) } \sqrt[4]{256}} & {\text { (e) } \sqrt[6]{\frac{1}{64}}}\end{array} $$

Short Answer

Expert verified
(a) \(\frac{2}{3}\), (b) 4, (e) \(\frac{1}{2}\).

Step by step solution

01

Solve (a)

The expression to evaluate is \( \sqrt{\frac{4}{9}} \). The square root of a fraction is the square root of the numerator divided by the square root of the denominator. Calculate: \( \sqrt{4} = 2 \) and \( \sqrt{9} = 3 \), so \( \sqrt{\frac{4}{9}} = \frac{2}{3} \).
02

Solve (b)

The expression to evaluate is \( \sqrt[4]{256} \). We need to find the number that, when raised to the fourth power, gives 256. By trial or calculation: \( 4^4 = 256 \). Thus, \( \sqrt[4]{256} = 4 \).
03

Solve (e)

The expression to evaluate is \( \sqrt[6]{\frac{1}{64}} \). The sixth root of a fraction is the sixth root of the numerator divided by the sixth root of the denominator. Calculate: \( \sqrt[6]{1} = 1 \) and \( \sqrt[6]{64} = 2 \) (since \( 2^6 = 64 \)). Therefore, \( \sqrt[6]{\frac{1}{64}} = \frac{1}{2} \).

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Key Concepts

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

Square Root
The square root is a mathematical concept used to find a number which, when multiplied by itself, results in a given number. It's the opposite of squaring a number. For instance, if you have the number 9, the square root is 3 because 3 times 3 equals 9. In mathematical notation, it is represented with the radical symbol \( \sqrt{} \). The operation of finding the square root of a number is straightforward.
  • Identify the number you want to find the square root of.
  • Find a number which, when multiplied by itself, equals the original number.
  • If the number isn't a perfect square, the square root can be approximated or expressed in simplest radical form.
In the expression \( \sqrt{\frac{4}{9}} \), you find the square root of the fraction by separately taking the square root of the numerator and denominator. So, \( \sqrt{4} = 2 \) and \( \sqrt{9} = 3 \), giving \( \sqrt{\frac{4}{9}} = \frac{2}{3} \). This keeps math simple and logical, especially when the numbers are perfect squares.
Fourth Root
The fourth root of a number is a value that, when raised to the fourth power, equals the original number. Think of it as the opposite of raising a number to the fourth power. This concept is often less familiar than the square root but follows a similar approach. It's written using the radical symbol with an index of 4, like this: \( \sqrt[4]{} \).
  • Focus on finding a number that, when multiplied by itself four times, equals the number you're taking the root of.
  • Use either factorization or a calculator to simplify finding large fourth roots.
  • When dealing with fractions, consider the numerator and denominator separately.
For example, with \( \sqrt[4]{256} \), you're finding a number "x" such that \( x^4 = 256 \). After testing or computation, you find that 4 is the correct answer because \( 4 \times 4 \times 4 \times 4 = 256 \). By understanding these processes, you can feel confident tackling fourth root problems.
Sixth Root
The sixth root is another extension of the root concept, where you're looking for a number that, when raised to the sixth power, gives you the original number. Sixth roots are applicable in various complex problems and are shown as \( \sqrt[6]{} \).
  • Like other roots, treat numerators and denominators separately if working with fractions.
  • Utilize factor analysis or technology to find sixth roots, especially without memorization of higher powers.
  • Simplification can often occur when roots don't result in whole numbers, requiring rational expressions.
In the case of \( \sqrt[6]{\frac{1}{64}} \), you find the sixth root separately for the "1" and the "64". Here, \( \sqrt[6]{1} = 1 \) because any number to the sixth power is still 1, and \( \sqrt[6]{64} = 2 \) since \( 2^6 = 64 \). Therefore, \( \sqrt[6]{\frac{1}{64}} = \frac{1}{2} \). Understanding this concept helps in dissecting complicated roots into more manageable parts.

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