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 118: 73

Answer

$v=-\frac{4}{5} , -2$

Work Step by Step

We simplify as follows: $(\frac{v}{v+1})^{2}+\frac{2v}{v+1}=8$ $\frac{v^{2}}{(v+1)^{2}}+\frac{2v(v+1)}{(v+1)^{2}}=\frac{8(v+1)^{2}}{(v+1)^{2}}$ $v^{2}+2v(v+1)=8(v+1)^{2}$ $3v^{2}+2v=8v^{2}+16v+8$ $5v^{2}+14v+8=0$ $(5v+4)(v+2)=0$ $v=-\frac{4}{5}, -2$
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