Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - 8.1 Exercises: 36

Answer

$t=\left\{ \dfrac{-2-2\sqrt{3}}{5},\dfrac{-2+2\sqrt{3}}{5} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (5t+2)^2=12 ,$ take the square root of both sides (Square Root Property) and simplify the radical. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} 5t+2=\pm\sqrt{12} .\end{array} Simplifying the radical and then using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 5t+2=\pm\sqrt{4\cdot3} \\\\ 5t+2=\pm\sqrt{(2)^2\cdot3} \\\\ 5t+2=\pm2\sqrt{3} \\\\ 5t=-2\pm2\sqrt{3} \\\\ t=\dfrac{-2\pm2\sqrt{3}}{5} .\end{array} The solutions are \begin{array}{l}\require{cancel} t=\dfrac{-2-2\sqrt{3}}{5} \\\\\text{OR}\\\\ t=\dfrac{-2+2\sqrt{3}}{5} .\end{array} Hence, $ t=\left\{ \dfrac{-2-2\sqrt{3}}{5},\dfrac{-2+2\sqrt{3}}{5} \right\} .$
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