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: 46

Answer

No solution.

Work Step by Step

We are given: $\sqrt{x+5}+2=\sqrt{x-1}$ We square both sides: $(\sqrt{x+5}+2)^{2}=(\sqrt{x-1})^{2}$ $(\sqrt{x+5}+2)(\sqrt{x+5}+2)=(\sqrt{x-1})(\sqrt{x-1})$ And distribute: $(x+5)+4\sqrt{x+5}+4=x-1$ And combine like terms: $9+4\sqrt{x+5}+1=x-x$ $4\sqrt{x+5}=-10$ $2\sqrt{x+5}=-5$ We square both sides again: $(2\sqrt{x+5})^{2}=(-5)^{2}$ And distribute: $4(x+5)=25$ $4x+20=25$ $4x=5$ $x=\frac{5}{4}$ However, the solution $\frac{5}{4}$ does not work in the original equation: $\sqrt{\frac{5}{4}+5}+2=\sqrt{\frac{5}{4}-1}$ $\sqrt{\frac{5}{4}+\frac{20}{4}}+2=\sqrt{\frac{5}{4}-\frac{4}{4}}$ $\sqrt{\frac{25}{4}}+2=\sqrt{\frac{1}{4}}$ $\frac{\sqrt{25}}{\sqrt{4}}+2=\frac{\sqrt{1}}{\sqrt{4}}$ $\frac{5}{2}+2=\frac{1}{2}$ $\frac{9}{2}=\frac{1}{2}$ $9=1$ Since we got a false statement, the solution $\frac{5}{4}$ does not work. Hence there is no solution.
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