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: 47


$x=0, 4$

Work Step by Step

$\sqrt {3x+4} -1=\sqrt {2x+1}$ $\sqrt {3x+4} -1+1=\sqrt {2x+1}+1$ $\sqrt {3x+4} =\sqrt {2x+1}+1$ $(\sqrt {3x+4})^2 =(\sqrt {2x+1}+1)^2$ $3x+4=(\sqrt {2x+1})(\sqrt {2x+1})+1*(\sqrt {2x+1})+1*(\sqrt {2x+1})+1*1$ $3x+4=2x+1+2\sqrt {2x+1}+1$ $3x+4=2x+2+2\sqrt {2x+1}$ $x+2=2\sqrt {2x+1}$ $(x+2)^2=(2\sqrt {2x+1})^2$ $x*x+2*x+x*2+2*2=2*2*(\sqrt{2x+1})^2$ $x^2+4x+4=4(2x+1)$ $x^2+4x+4=8x+4$ $x^2-4x=0$ $x(x-4)=0$ $x=0$ $x-4=0$ $x=4$ $\sqrt {3x+4} -1=\sqrt {2x+1}$ $\sqrt {3*0+4} -1=\sqrt {2*0+1}$ $\sqrt {4} -1=\sqrt {1}$ $2-1=1$ (true) $\sqrt {3x+4} -1=\sqrt {2x+1}$ $\sqrt {3*4+4} -1=\sqrt {2*4+1}$ $\sqrt {12+4} -1=\sqrt {8+1}$ $\sqrt {16} -1=\sqrt {9}$ $4-1=3$ (true)
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