College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises - Page 135: 52

Answer

$x=27$

Work Step by Step

We are given: $\sqrt{2x-5}=2+\sqrt{x-2}$ We square both sides: $(\sqrt{2x-5})^{2}=(2+\sqrt{x-2})^{2}$ $(\sqrt{2x-5})(\sqrt{2x-5})=(2+\sqrt{x-2})(2+\sqrt{x-2})$ $2x-5=4+4\sqrt{x-2}+x-2$ $2x-5-2=x+4\sqrt{x-2}$ $x-7=4\sqrt{x-2}$ We square both sides again: $(x-7)^{2}=(4\sqrt{x-2})^{2}$ $x^{2}-14x+49=16(x-2)$ And distribute: $x^{2}-14x+49=16x-32$ $x^{2}-30x+81=0$ And factor: $(x-3)(x-27)=0$ Use the zero-factor property by equating each factor to zero: $(x-3)=0$ or $(x-27)=0$ $x=3$ or $x=27$ However, the solution $x=3$ does not work in the original equation: $\sqrt{2(3)-5}=2+\sqrt{3-2}$ $\sqrt{6-5}=2+\sqrt{1}$ $\sqrt{1}=2+1$ $1=3$ Since we got a false statement, $x=3$ is not a solution. Thus the only solution is $x=27$.
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