College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 179: 88


$x=6$ satisfies the given conditions

Work Step by Step

$y=x-\sqrt{x-2}$ and $y=4$ Substitute $y$ by $4$: $x-\sqrt{x-2}=4$ Take $\sqrt{x-2}$ to the right side and $4$ to the left side: $x-4=\sqrt{x-2}$ Square both sides of the equation: $(x-4)^{2}=(\sqrt{x-2})^{2}$ $x^{2}-8x+16=x-2$ Take all terms to the left side of the equation: $x^{2}-8x+16-x+2=0$ Simplify: $x^{2}-9x+18=0$ Solve by factoring: $(x-6)(x-3)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-6=0$ $x=6$ $x-3=0$ $x=3$ Check the solutions found by plugging them into the original equation: $x=3$ $3-\sqrt{3-2}=4$ False $x=6$ $6-\sqrt{6-2}=4$ True $x=6$ satisfies the given conditions
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