Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.6 - Radical Equations and Problem Solving - Exercise Set - Page 729: 49



Work Step by Step

$\sqrt {y+3}-\sqrt{y-3}=1$ $\sqrt {y+3}-\sqrt{y-3}+\sqrt{y-3}=1+\sqrt{y-3}$ $\sqrt {y+3}=1+\sqrt{y-3}$ $(\sqrt {y+3})^2=(1+\sqrt{y-3})^2$ $(y+3)=(1+\sqrt{y-3})^2$ $y+3=1*1+(1*\sqrt{y-3})+(1*\sqrt{y-3})+(\sqrt{y-3})(\sqrt{y-3})$ $y+3=1+2\sqrt{y-3}+(y-3)$ $y+3=1+2\sqrt{y-3}+y-3$ $y+3=-2+2\sqrt{y-3}+y$ $3=-2+2\sqrt{y-3}$ $3+2=-2+2\sqrt{y-3}+2$ $5=2\sqrt{y-3}$ $5/2=2\sqrt{y-3}/2$ $5/2=\sqrt{y-3}$ $(5/2)^2=(\sqrt{y-3})^2$ $25/4=y-3$ $25/4+3=y-3+3$ $25/4+12/4=y$ $37/4=y$ $\sqrt {y+3}-\sqrt{y-3}=1$ $\sqrt {37/4+3}-\sqrt{37/4-3}=1$ $\sqrt {37/4+12/4}-\sqrt{37/4-12/4}=1$ $\sqrt {49/4}-\sqrt{25/4}=1$ $7/2-5/2=1$ $2/2=1$ (true)
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