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 - Page 113: 76



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of equality to solve the given equation, $ s=s_0+gt^2+k ,$ for $ t .$ $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} s-s_0-k=gt^2 \\\\ \dfrac{s-s_0-k}{g}=t^2 \\\\ t^2=\dfrac{s-s_0-k}{g} .\end{array} Taking the square root of both sides (Square Root Principle) results to \begin{array}{l}\require{cancel} t=\pm\sqrt{\dfrac{s-s_0-k}{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{s-s_0-k}{g}\cdot\dfrac{g}{g}} \\\\ t=\pm\sqrt{\dfrac{g(s-s_0-k)}{g^2}} \\\\ t=\pm\sqrt{\dfrac{1}{g^2}\cdot g(s-s_0-k)} \\\\ t=\pm\sqrt{\left( \dfrac{1}{g} \right)^2 \cdot g(s-s_0-k)} \\\\ t=\pm\left| \dfrac{1}{g} \right|\sqrt{g(s-s_0-k)} .\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{g(s-s_0-k)} \\\\ t=\pm\dfrac{\sqrt{g(s-s_0-k)}}{g} .\end{array}
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