Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set: 14

Answer

$x=\pm4$

Work Step by Step

Using the properties of equality, the given equation, $ 3x^2-1=47 ,$ is equivalent to \begin{array}{l}\require{cancel} 3x^2=47+1 \\\\ 3x^2=48 \\\\ x^2=\dfrac{48}{3} \\\\ x^2=16 .\end{array} Taking the square root of both sides, then the solutions of the given equation are \begin{array}{l}\require{cancel} x=\pm\sqrt{16} \\\\ x=\pm\sqrt{(4)^2} \\\\ x=\pm4 .\end{array}
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