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: 51

Answer

$x=-2$

Work Step by Step

We are given: $\sqrt{x+2}=1-\sqrt{3x+7}$ We square both sides: $(\sqrt{x+2})^{2}=(1-\sqrt{3x+7})^{2}$ $(\sqrt{x+2})(\sqrt{x+2})=(1-\sqrt{3x+7})(1-\sqrt{3x+7})$ $x+2=1-2\sqrt{3x+7}+(3x+7)$ $x+2-3x-8=-2\sqrt{3x+7}$ $-2x-6=-2\sqrt{3x+7}$ $2\sqrt{3x+7}=2x+6$ $2\sqrt{3x+7}=2(x+3)$ $\sqrt{3x+7}=x+3$ We square both sides again: $(\sqrt{3x+7})^{2}=(x+3)^{2}$ $3x+7=x^{2}+6x+9$ $x^{2}+3x+2=0$ And factor: $(x+2)(x+1)=0$ Use the zero-factor property by equating each factor to zero: $(x+2)=0$ or $(x+1)=0$ $x=-2$ or $x=-1$ However, the solution $x=-1$ does not work in the original equation: $\sqrt{-1+2}=1-\sqrt{3*-1+7}$ $\sqrt{1}=1-\sqrt{-3+7}$ $1=1-\sqrt{4}$ $1=1-2$ $1=-1$ Since we got a false statement, the solution $x=-1$ does not work. Thus the only solution is $x=-2$.
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