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 - Page 536: 7


$t=\pm\dfrac{\sqrt{dk }}{k}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ d=kt^2 ,$ in terms of $ t ,$ use the properties of equality and the Square Root Principle to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{d}{k}=t^2 \\\\ t^2=\dfrac{d}{k} .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} t=\pm\sqrt{\dfrac{d}{k}} .\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} t=\pm\sqrt{\dfrac{d}{k}\cdot\dfrac{k}{k}} \\\\ t=\pm\sqrt{\dfrac{dk}{k^2}} \\\\ t=\pm\sqrt{\dfrac{1}{k^2}\cdot dk } \\\\ t=\pm\sqrt{\left(\dfrac{1}{k} \right)^2\cdot dk } \\\\ t=\pm\dfrac{1}{k}\sqrt{dk } \\\\ t=\pm\dfrac{\sqrt{dk }}{k} .\end{array}
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