/*! 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 16 In Exercises \(1-21,\) solve the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(1-21,\) solve the equation for the variable. $$ \frac{1}{4} t^{3}=\frac{4}{t} $$

Short Answer

Expert verified
Question: Solve the equation \(\frac{1}{4}t^3 = \frac{4}{t}\) for variable \(t\). Answer: The solutions for the given equation are \(t = 2\) and \(t = -2\).

Step by step solution

01

Eliminate the fractions

To eliminate the fractions, we need to find the LCM, which in this case is \(4t\). Multiply both sides of the equation by \(4t\) to eliminate the fractions: $$ (4t) \left( \frac{1}{4} t^{3} \right) = (4t) \left( \frac{4}{t} \right) $$
02

Simplify the equation

Simplify both sides of the equation after multiplying with LCM: $$ t^{4} = 16 $$
03

Isolate the variable t

We have a simple polynomial equation \(t^4 = 16\). To solve for t, we need to take the fourth root on both sides of the equation: $$ t = \pm \sqrt[4]{16} $$
04

Calculate the solution(s) for t

Find the fourth root of 16 and consider both positive and negative values as solutions: $$ t = \pm 2 $$ So the solutions for the given equation are \(t = 2\) and \(t = -2\).

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 91Ó°ÊÓ!

Key Concepts

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

Fraction Elimination in Equations
Eliminating fractions in equations is a fundamental skill in algebra. The presence of fractions can make solving equations more cumbersome. Thus, the first step is to eliminate them. You do this by multiplying each term of the equation by the least common multiple (LCM) of the denominators that appear in the equation.
For instance, in the equation \( \frac{1}{4} t^{3} = \frac{4}{t} \), the LCM of the denominators \( 4 \) and \( t \) is \( 4t \). By multiplying each side by \( 4t \), you remove the fractions from the equation:
  • Multiply \( (4t) \left( \frac{1}{4} t^{3} \right) \) to get \( t^4 \)
  • Multiply \( (4t) \left( \frac{4}{t} \right) \) to get \( 16 \)
This results in the simplified equation \( t^4 = 16 \). Eliminating fractions simplifies solving equations, making it a vital skill to master.
Understanding Polynomial Equations
Polynomial equations are equations involving a polynomial expression in terms of a variable. A polynomial is a sum of terms with variables raised to whole number powers and coefficients. In this context, \( t^4 = 16 \) is a polynomial equation where the highest power of \( t \) is 4.
Polynomial equations are named by their degree, which is the highest exponent of the variable. Here, it's called a "quartic" equation because of its degree of 4.
Solving polynomial equations often requires isolating the variable. This might involve factoring, using the quadratic formula, or performing operations like taking roots. In this example, you isolate \( t \) by computing the fourth root on both sides:
  • \( t^4 = 16 \)
  • \( t = \pm \sqrt[4]{16} \)
Understanding polynomials and how to solve them will equip you with techniques to tackle more complex algebraic problems.
Finding the Roots of Equations
The roots of an equation are the solutions that make the equation true. They can be interpreted as the x-values where the graph of a polynomial crosses the x-axis. To find the roots, you solve for the variable.
In the equation \( t^4 = 16 \), you look for values of \( t \) that satisfy the equation. This process often involves extracting roots. Here, it requires taking the fourth root of both sides. Remember that both positive and negative values are valid for even roots:
  • The fourth root of 16 is 2.
  • Thus, \( t = 2 \) and \( t = -2 \) are both roots of the given equation.
These are real numbers indicating the points where the polynomial equation equals zero. Finding roots is essential as it helps solve equations by pinpointing their 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

When \(P\) dollars is invested at an annual interest rate \(r\) compounded once a year, the balance, \(A,\) after 2 years is given by \(A=P(1+r)^{2}\). (a) Evaluate \(A\) when \(r=0,\) and interpret the answer in practical terms. (b) If \(r\) is between \(5 \%\) and \(6 \%,\) what can you conclude about the percentage growth after 2 years? (c) Express \(r\) as a function of \(A\). Under what circumstances might this function be useful? (d) What interest rate is necessary to obtain an increase of \(25 \%\) in 2 years?

Which of the following steps is the appropriate next step to solve the equation \(x^{3}+8=64\) ? (a) Take the cube root of both sides of the equation (b) Subtract 8 from both sides of the equation.

In Problems \(22-25,\) find possible formulas for the power functions. $$ \begin{array}{l|l|c|c|c} \hline x & 1 & 2 & 3 & 4 \\ \hline y & 4 & 16 & 36 & 64 \\ \hline \end{array} $$

Poiseuille's Law tells us that the rate of flow, \(R,\) of a gas through a cylindrical pipe is proportional to the fourth power of the radius, \(r\), of the pipe, given a fixed drop in pressure between the two ends of the pipe. For a certain gas, if the rate of flow is measured in \(\mathrm{cm}^{3} / \mathrm{sec}\) and the radius is measured in \(\mathrm{cm}\), the constant of proportionality is 4.94 . (a) If the rate of flow of this gas through a pipe is 500 \(\mathrm{cm}^{3} / \mathrm{sec},\) what is the radius of the pipe? (b) Solve for the radius \(r\) in terms of the rate of flow \(R\). (c) Is \(r\) proportional to a power of \(R ?\) If so, what power?

In Problems \(59-70,\) decide for what values of the constant \(A\) the equation has (a) The solution \(t=0\) (b) A positive solution (c) A negative solution $$ t^{3}=A^{2} $$

See all solutions

Recommended explanations on Math Textbooks

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.