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: 48


$x= 2,6$

Work Step by Step

$\sqrt {x-2} +3=\sqrt {4x+1}$ $\sqrt {x-2} +3-3=\sqrt {4x+1}-3$ $\sqrt {x-2} =\sqrt {4x+1}-3$ $(\sqrt {x-2})^2 =(\sqrt {4x+1}-3)^2$ $x-2=(\sqrt {4x+1})*(\sqrt {4x+1})+(\sqrt {4x+1})*(-3)+(-3)(\sqrt {4x+1})+(-3)(-3)$ $x-2=(4x+1)-6\sqrt {4x+1}+9$ $x-2=4x+10-6\sqrt {4x+1}$ $-3x-2=10-6\sqrt {4x+1}$ $-3x-12=-6\sqrt {4x+1}$ $(-3x-12)^2=(-6\sqrt {4x+1})^2$ $(-3x)^2+(-3x)(-12)+(-12)(-3x)+(-12)(-12)=(-6)^2(\sqrt {4x+1})^2$ $9x^2+72x+144=36(4x+1)$ $9x^2+72x+144=144x+36$ $9x^2-72x+144=36$ $9x^2-72x+108=0$ $(9x^2-72x+108=0)/9$ $x^2-8x+12=0$ $(x-2)(x-6)=0$ $x-2=0$ $x=2$ $x-6=0$ $x=6$ $\sqrt {x-2} +3=\sqrt {4x+1}$ $\sqrt {2-2} +3=\sqrt {4*2+1}$ $\sqrt {0} +3=\sqrt {8+1}$ $0+3=\sqrt {9}$ $0+3=3$ $3=3$ (true) $\sqrt {x-2} +3=\sqrt {4x+1}$ $\sqrt {6-2} +3=\sqrt {4*6+1}$ $\sqrt {4} +3=\sqrt {24+1}$ $2 +3=\sqrt {25}$ $5=5$ (true)
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