Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set: 15

Answer

$t=\pm\dfrac{\sqrt{30}}{6}$

Work Step by Step

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $ 6t^2-5=0 ,$ are \begin{array}{l}\require{cancel} 6t^2=5 \\\\ t^2=\dfrac{5}{6} \\\\ t=\pm\sqrt{\dfrac{5}{6}} \\\\ t=\pm\sqrt{\dfrac{5}{6}\cdot\dfrac{6}{6}} \\\\ t=\pm\sqrt{\dfrac{30}{36}} \\\\ t=\pm\dfrac{\sqrt{30}}{\sqrt{36}} \\\\ t=\pm\dfrac{\sqrt{30}}{\sqrt{(6)^2}} \\\\ t=\pm\dfrac{\sqrt{30}}{6} .\end{array}
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