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: 40

Answer

$y=\left\{ 7-3\sqrt{2},7+3\sqrt{2} \right\}$

Work Step by Step

The left side of the given equation, $ y^2-14y+49=18 ,$ is a perfect square trinomial whose factored form is \begin{array}{l}\require{cancel} (y-7)^2=18 .\end{array} Taking the square root of both sides (the Square Root Property), the solutions to the given equation are \begin{array}{l}\require{cancel} y-7=\pm\sqrt{18} \\\\ y-7=\pm\sqrt{9\cdot2} \\\\ y-7=\pm\sqrt{(3)^2\cdot2} \\\\ y-7=\pm3\sqrt{2} \\\\ y=7\pm3\sqrt{2} .\end{array} Hence, $ y=\left\{ 7-3\sqrt{2},7+3\sqrt{2} \right\} .$
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