## College Algebra 7th Edition

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