College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 117: 30

Answer

Solution set: $\{-2\}$

Work Step by Step

$\sqrt{3x+7}+\sqrt{x+2}=1$ Subtract $\sqrt{x+2}$ from both sides $\sqrt{3x+7}=1-\sqrt{x+2}$ Square both sides $(\sqrt{3x+7})^{2}=(1-\sqrt{x+2})^{2}$...Apply $(a-b)^{2}=a^{2}-2ab+b^{2}$ $3x+7=1-2\sqrt{x+2}+(x+2)$ $2x+4=-2\sqrt{x+2}$ Square both sides $4x^{2}+16x+16=4(x+2)$ $4x^{2}+16x+16=4x+8$ $4x^{2}+12x+8=0$ $x^{2}+3x+2=0$ $(x+1)(x+2)=0$ Testing $x=-1,$ $\sqrt{3(-1)+7}+\sqrt{(-1)+2}=3+1=4$ ... not a solution. Testing $x=-2,$ $\sqrt{3(-2)+7}+\sqrt{(-2)+2}=0+1=1$ ... $-2$ is a solution. Solution set: $\{-2\}$
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