Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 148: 60

Answer

$\{8\}$

Work Step by Step

Step 1. Rewrite the equation as $\sqrt {2x}+2=\sqrt {3x+12}$ Step 2. Take square on both sides to get: $2x+4\sqrt {2x}+4=3x+12$ or $4\sqrt {2x}=x+8$ Step 3. Take square on both sides again to get: $32x=x^2+16x+64$ or $x^2-16x+64=0$ which gives $x=8$ Step 4. Check answer in the original equation: it fits.
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