College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises: 48

Answer

$x=\frac{-1\pm\sqrt{5}}{2}$

Work Step by Step

We need to solve: $x^{2}e^{x}+xe^{x}-e^{x}=0$ We factor: $e^{x}(x^{2}+x-1)=0$ We have either: $e^{x}=0$ $x=\ln 0$=no solution Or: $x^{2}+x-1=0$ $4x^{2}+4x-4=0$ $4x^{2}+4x+1-5=0$ $(2x+1)^2-5=0$ $(2x+1)^2=5$ $2x+1=\pm\sqrt{5}$ $2x=\pm\sqrt{5}-1$ $x=\frac{-1\pm\sqrt{5}}{2}$ (We could also have used the quadratic formula.)
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