College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises: 71

Answer

$t=\pm\dfrac{\sqrt{2sg}}{g}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of equality to solve the given equation, $ s=\dfrac{1}{2}gt^2 ,$ for $ t .$ $\bf{\text{Solution Details:}}$ Multiplying both sides by $ 2 ,$ and then dividing by $ g $ result to \begin{array}{l}\require{cancel} 2s=gt^2 \\\\ \dfrac{2s}{g}=t^2 \\\\ t^2=\dfrac{2s}{g} .\end{array} Taking the square root of both sides (Square Root Principle) results to \begin{array}{l}\require{cancel} t=\pm\sqrt{\dfrac{2s}{g}} .\end{array} 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{2s}{g}\cdot\dfrac{g}{g}} \\\\ t=\pm\sqrt{\dfrac{2sg}{g^2}} \\\\ t=\pm\sqrt{\dfrac{1}{g^2}\cdot2sg} \\\\ t=\pm\sqrt{\left( \dfrac{1}{g} \right)^2\cdot2sg} \\\\ t=\pm\left| \dfrac{1}{g} \right|\sqrt{2sg} .\end{array} Assuming that all variables are positive, the equation above is equivalent to \begin{array}{l}\require{cancel} t=\pm\dfrac{1}{g}\sqrt{2sg} \\\\ t=\pm\dfrac{\sqrt{2sg}}{g} .\end{array}
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