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Chapter 10: Problem 55

In Exercises \(39-64,\) rationalize each denominator. $$\frac{7}{\sqrt[3]{2 x^{2}}}$$

Short Answer

Expert verified
The fraction \(\frac{7}{\sqrt[3]{2 x^{2}}}\) with the denominator rationalized is \(\frac{7\sqrt[3]{(2x)}}{2x}\).

Step by step solution

01

Identify the denominator

First, the denominator is identified: \(\sqrt[3]{2 x^{2}}\). In order to rationalize this denominator, it should be multiplied by something that would make the total power equal to 3 (because of the cube root). Hence the need to multiply by a form of 1, \(\sqrt[3]{(2x)}\) over itself because \(2x^{2}\) * \(2x\) would give \(4x^3\) which is a whole number under cuberoot.
02

Multiply the fraction by the form of 1

Multiply the original fraction by the form of 1, which gives: \(\frac{7}{\sqrt[3]{2 x^{2}}}\) * \(\frac{\sqrt[3]{(2x)}}{\sqrt[3]{(2x)}}\). This gives you: \(\frac{7\sqrt[3]{(2x)}}{2x}\)
03

Simplify if possible

The fraction can't be simplified further. Therefore, the fraction with the denominator rationalized is \(\frac{7\sqrt[3]{(2x)}}{2x}\).

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Key Concepts

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

Cube Roots
Cube roots are essential when rationalizing denominators, especially in expressions involving roots. A cube root, denoted as \( \sqrt[3]{a} \), is a number that when multiplied by itself three times, equals \( a \). In our exercise, we have the cube root \( \sqrt[3]{2x^2} \). To rationalize this, our goal is to "clear" the root from the denominator and transform the expression into a simpler form.
  • Cube roots involve recognizing numbers that need to be multiplied to themselves three times to form another number.
  • Rationalizing requires finding a way to multiply these roots to obtain a simple number, in this case by making the power of the variables a multiple of three.
This process involves strategic multiplication, allowing us to work with a clearer algebraic expression free of cube roots in the denominator.
Multiplying Fractions
Understanding how to multiply fractions correctly is key to rationalizing and simplifying expressions like the one in the exercise. When multiplying fractions, you multiply the numerators together and then multiply the denominators together. The original expression \( \frac{7}{\sqrt[3]{2 x^{2}}} \) needs to be multiplied by \( \frac{\sqrt[3]{(2x)}}{\sqrt[3]{(2x)}} \), which is a form of 1.
  • Multiplying by 1 does not change the value of the fraction but allows us to manipulate the form.
  • By carefully selecting \( \sqrt[3]{(2x)} \), we ensure that the denominator becomes \( 2x \), a non-root term.
This strategic multiplication helps us "rationalize", letting us rewrite the original fraction in a form that鈥檚 easier to work with without altering its value.
Simplification of Algebraic Expressions
Simplification involves reducing expressions to their simplest or most convenient form. After rationalizing the denominator, the expression may be written as \( \frac{7\sqrt[3]{(2x)}}{2x} \). At this stage, checking for any further simplification is essential.
  • Simplification can often involve factorization or cancellation if common terms exist.
  • In this case, the fraction has been reduced as far as possible; no further simplification is available.
Simplification aids in reducing computation complexity, making mathematical expressions more manageable and easier to understand across various contexts.

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Most popular questions from this chapter

In Exercises \(75-92,\) rationalize each denominator. Simplify, if possible. $$\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}$$

In Exercises \(65-74,\) simplify each radical expression and then rationalize the denominator. $$-\sqrt{\frac{75 a^{5}}{b^{3}}}$$

In Exercises \(93-104\), rationalize each numerator. Simplify, if possible. $$\sqrt{\frac{5}{3}}$$

In Exercises \(65-74,\) simplify each radical expression and then rationalize the denominator. $$\frac{25}{\sqrt{5 x^{2} y}}$$

In Exercises \(105-112,\) add or subtract as indicated. Begin by rationalizing denominators for all terms in which denominators contain radicals. $$\sqrt{2}+\frac{1}{\sqrt{2}}$$
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