Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.4 - Formulas and Further Applications - 8.4 Exercises: 12

Answer

$d=\pm\dfrac{\sqrt{kR}}{R}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ R=\dfrac{k}{d^2} ,$ in terms of $ d ,$ use the properties of equality and the Square Root Principle to isolate the variable. $\bf{\text{Solution Details:}}$ Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} R(d^2)=1(k) \\\\ Rd^2=k \\\\ d^2=\dfrac{k}{R} .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} d=\pm\sqrt{\dfrac{k}{R}} .\end{array} Rationalizing the denominator by multiplying the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} d=\pm\sqrt{\dfrac{k}{R}\cdot\dfrac{R}{R}} \\\\ d=\pm\sqrt{\dfrac{kR}{R^2}} \\\\ d=\pm\sqrt{\dfrac{1}{R^2}\cdot kR} \\\\ d=\pm\sqrt{\left( \dfrac{1}{R} \right)^2\cdot kR} \\\\ d=\pm\dfrac{1}{R}\sqrt{kR} \\\\ d=\pm\dfrac{\sqrt{kR}}{R} .\end{array}
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