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: 28

Answer

$t=\dfrac{-3\pm\sqrt{14}}{2}$

Work Step by Step

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), then the solutions to the given equation, $ \left( t+\dfrac{3}{2} \right)^2=\dfrac{7}{2} ,$ are \begin{array}{l}\require{cancel} t+\dfrac{3}{2}=\pm\sqrt{\dfrac{7}{2}} \\\\ t+\dfrac{3}{2}=\pm\sqrt{\dfrac{7}{2}\cdot\dfrac{2}{2}} \\\\ t+\dfrac{3}{2}=\pm\sqrt{\dfrac{14}{4}} \\\\ t+\dfrac{3}{2}=\pm\dfrac{\sqrt{14}}{\sqrt{4}} \\\\ t+\dfrac{3}{2}=\pm\dfrac{\sqrt{14}}{(2)^2} \\\\ t+\dfrac{3}{2}=\pm\dfrac{\sqrt{14}}{2} \\\\ t=-\dfrac{3}{2}\pm\dfrac{\sqrt{14}}{2} \\\\ t=\dfrac{-3\pm\sqrt{14}}{2} .\end{array}
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