/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Simplify the expression, and rat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 10

Simplify the expression, and rationalize the denominator when appropriate. $$\sqrt[4]{\left(-4 a^{3} b^{2} c\right)^{2}}$$

Short Answer

Expert verified
The simplified expression is \(2a^{3/2}b c^{1/2}\).

Step by step solution

01

Apply the Power Rule

The expression \((-4a^3b^2c)^2\) can be simplified using the power rule: \[(xy)^n = x^n \cdot y^n\].Applying it here, we have: \[(-4)^2 \cdot (a^3)^2 \cdot (b^2)^2 \cdot c^2\].This results in the expression: \[16 \cdot a^6 \cdot b^4 \cdot c^2\].
02

Apply the Fourth Root

Now we simplify \(\sqrt[4]{16 \cdot a^6 \cdot b^4 \cdot c^2}\).The fourth root can be distributed as:\[\sqrt[4]{16} \cdot \sqrt[4]{a^6} \cdot \sqrt[4]{b^4} \cdot \sqrt[4]{c^2}\].
03

Simplify Each Component

Let's break down each component:- \(\sqrt[4]{16}\) simplifies to 2, since \(2^4 = 16\).- \(\sqrt[4]{a^6}\) can be rewritten as \((a^6)^{1/4} = a^{6/4} = a^{3/2}\).- \(\sqrt[4]{b^4}\) simplifies directly to \(b\).^4- \(\sqrt[4]{c^2}\) can be rewritten as \((c^2)^{1/4} = c^{1/2}\).
04

Write the Simplified Expression

Putting the simplified components back together, we get:\[2a^{3/2}b c^{1/2}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Power Rule
When simplifying expressions involving exponents, the power rule is a handy tool. It tells us that
  • \[(xy)^n = x^n \cdot y^n\]
  • Meaning, when raising a product to a power, apply the exponent to each factor separately.
  • This rule helps to break down complex expressions into more manageable parts, making computations easier.
In our problem, we initially have the expression \((-4a^3b^2c)^2\). Applying the power rule, each component within the bracket is squared:\[(-4)^2 \cdot (a^3)^2 \cdot (b^2)^2 \cdot c^2\].This gives us:
  • \[16 \cdot a^6 \cdot b^4 \cdot c^2\]
Thus, the power rule simplifies our task by applying the power evenly across all elements.
Fourth Root
The concept of a fourth root is quite straightforward but essential in various algebraic manipulations.
  • A fourth root of a number \(x\) is a number that, when raised to the power of four, equals \(x\).
  • You represent it as \(\sqrt[4]{x}\).
Applying this to the expression \(\sqrt[4]{16 \cdot a^6 \cdot b^4 \cdot c^2}\), we deal with each element separately:
  • The fourth root of 16 is 2, since \(2^4 = 16\).
  • For \(a^6\), the fourth root involves managing exponents, specifically \[(a^6)^{1/4} = a^{6/4} = a^{3/2}\].
  • For \(b^4\), the fourth root simplifies neatly to \(b\).
  • For \(c^2\), consider \[(c^2)^{1/4} = c^{1/2}\].
Each element is simplified by applying the concept of fractional exponents, which is integral when dealing with roots.
Simplifying Expressions
Simplifying expressions is a fundamental aspect of algebra. It involves:
  • Breaking down complicated expressions into simpler, more manageable parts.
  • Ensuring each expression is written in its simplest form.
  • This often requires operations like distributing powers, combining like terms, or factoring where possible.
In this exercise, after applying the fourth root to each factor, we simplify further to achieve clarity and simplicity. Expressions such as \(a^{6/4}\) are reduced to \(a^{3/2}\) by simplifying the fraction.
  • This results in the expression \[2a^{3/2}b c^{1/2}\], which is a simpler, more understandable version of the original complex expression.
Algebraic Manipulation
Algebraic manipulation allows us to transform expressions to suit different needs, such as simplifying, solving, or reformatting.
  • The objective is often to make an equation or expression easier to understand or solve.
  • Common techniques include using identities, removing parentheses, or distributing powers.
  • In this exercise, we used several forms of manipulation: applying the power rule, finding roots, and simplifying the expression.
Moreover, algebraic manipulation can involve rationalizing denominators, especially when dealing with radicals. Although not needed here specifically, it's critical to understand that it's about removing any radicals from the denominator for a polished and conventional form. Ultimately, mastering manipulation helps in various mathematical tasks, ensuring precision and clarity in solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Apartment rentals \(A\) real estate company owns 218 efficiency apartments, which are fully occupied when the rent is 940 dollars per month. The company estimates that for each 25 dollars increase in rent, 5 apartments will become unoccupied. What rent should be charged in order to pay the monthly bills, which total 205,920 dollars ?

Simplify the expression, assuming \(x\) and \(y\) may be negative. $$\sqrt[4]{(x+2)^{12} y^{4}}$$

Speed of sound The speed of sound in air at \(0^{\circ} \mathrm{C}(\text { or } 273 \mathrm{K})\) is \(1087 \mathrm{ft} / \mathrm{sec},\) but this speed increases as the temperature rises. The speed \(v\) of sound at temperature \(T\) in \(K\) is given by \(v=1087 \sqrt{T / 273} .\) At what temperatures does the speed of sound exceed \(1100 \mathrm{ft} / \mathrm{sec} ?\)

Every cross section of a drainage ditch is an isosceles trapezoid with a small base of 3 feet and a height of 1 foot, as shown in the figure. Determine the width of the larger base that would give the ditch a cross-sectional area of \(5 \mathrm{ft}^{2}\).

A large grain silo is to be constructed in the shape of a circular cylinder with a hemisphere attached to the top (see the figure). The diameter of the silo is to be 30 feet, but the height is yet to be determined. Find the height \(h\) of the silo that will result in a capacity of \(11,250 \pi \mathrm{ft}^{3}\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.