Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.6 - Radical Equations and Problem Solving - Exercise Set: 19

Answer

no solution

Work Step by Step

Squaring both sides of the given equation, $ \sqrt{y+5}=2-\sqrt{y-4} ,$ results to \begin{array}{l}\require{cancel} (\sqrt{y+5})^2=(2-\sqrt{y-4})^2 \\ y+5=(2)^2-2(2)(\sqrt{y-4})+(\sqrt{y-4})^2 \\ y+5=4-4\sqrt{y-4}+y-4 \\ y+5=y-4\sqrt{y-4} \\ y-y+5=-4\sqrt{y-4} \\ 5=-4\sqrt{y-4} \\ \dfrac{5}{-4}=\sqrt{y-4} \\ \sqrt{y-4}=-\dfrac{5}{4} .\end{array} The radical expression at the left side of the equation above reults to a nonnegative number. This will never be equal to the negative expression at the right. Hence, there is $\text{ no solution }.$
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