Intermediate Algebra (6th Edition)

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

Chapter 8 - Cumulative Review - Page 534: 44

Answer

$x=6$, $x=2$

Work Step by Step

$\sqrt {x-2} = \sqrt {4x+1} -3$ $(\sqrt {x-2})^2 = (\sqrt {4x+1} -3)^2$ $x-2 = (\sqrt {4x+1}-3)*(\sqrt {4x+1}-3)$ $x-2 = \sqrt {4x+1}*\sqrt {4x+1}+(-3)(\sqrt {4x+1})+(-3)(\sqrt {4x+1})+(-3)*(-3)$ $x-2 = (4x+1) -6\sqrt {4x+1} + 9$ $x-2=4x+1+9-6\sqrt {4x+1}$ $x-2=4x+10-6\sqrt {4x+1}$ $x-2-4x-10=4x+10-6\sqrt {4x+1}-4x-10$ $-3x-12=-6\sqrt {4x+1}$ $(-3x-12)^2=(-6\sqrt {4x+1})^2$ $(-3x-12)*(-3x-12)=36*(4x+1)$ $(-3x)(-3x)+(-3x)(-12)+(-3x)(-12)+(-12)(-12) =36*4x+36$ $9x^2+36x+36x+144=144x+36$ $9x^2+72x+144=144x+36$ $9x^2+72x+144-144x-36=144x+36-144x-36$ $9x^2-72x+108=0$ $(9x^2-72x+108)/9=0/9$ $x^2-8x+12=0$ $(x-6)(x-2)=0$ $x-6=0$ $x-6+6=0+6$ $x=6$ $x-2=0$ $x-2+2=0+2$ $x=2$
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