College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 139: 73

Answer

The solution is $x=2$

Work Step by Step

$\sqrt{x+\sqrt{x+2}}=2\hspace{0.7cm}{\color{blue}{\text{Given equation}}}$ $\Rightarrow (\sqrt{x+\sqrt{x+2}})^2=2^2$ $\Rightarrow x+\sqrt{x+2}=4$ $\Rightarrow \sqrt{x+2}=4-x$ $\Rightarrow (\sqrt{x+2})^2=(4-x)^2$ $\Rightarrow x+2=16-8x+x^2$ $\Rightarrow x^2-9x+14=0$ $\Rightarrow (x-7)(x-2)=0$ $\Rightarrow x-7=0$ or $x-2=0$ $\Rightarrow x=7$ or $x=2$ $\underline{\textbf{Check the answer}}$ At $x=7$ LHS $\sqrt{7+\sqrt{7+2}}=\sqrt{7+\sqrt{9}}=\sqrt{7+3}=\sqrt{10}\ne$RHS $\hspace{0.3cm}{\color{red}{Reject}}$ At $x=2$ $\sqrt{2+\sqrt{2+2}}=\sqrt{2+\sqrt{4}}=\sqrt{2+2}=\sqrt{4}=2=$RHS The solution is $x=2$
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