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: 14

Answer

$h=\pm\dfrac{d^2\sqrt{kL}}{L}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ L=\dfrac{kd^4}{h^2} ,$ in terms of $ h ,$ 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} L(h^2)=1(kd^4) \\\\ Lh^2=kd^4 \\\\ h^2=\dfrac{kd^4}{L} .\end{array} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} h=\pm\sqrt{\dfrac{kd^4}{L}} .\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} h=\pm\sqrt{\dfrac{kd^4}{L}\cdot\dfrac{L}{L}} \\\\ h=\pm\sqrt{\dfrac{kd^4L}{L^2}} \\\\ h=\pm\sqrt{\dfrac{d^4}{L^2}\cdot kL} \\\\ h=\pm\sqrt{\left( \dfrac{d^2}{L} \right)^2\cdot kL} \\\\ h=\pm\dfrac{d^2}{L}\sqrt{kL} \\\\ h=\pm\dfrac{d^2\sqrt{kL}}{L} .\end{array}
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