The equation has two roots. Find the smaller and larger one? so I came up with 2x^{2}e^{-5x}(3-3x)=0 Just not sure where to go from here?
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Originally Posted by M670 The equation has two roots. Find the smaller and larger one? so I came up with 2x^{2}e^{-5x}(3-3x)=0 Just not sure where to go from here? You need to use the zero product principle. This would give $\displaystyle 2x^2=0 \quad e^{-5x}=0 \quad 3-3x=0$ Only two of these equations have solutions.
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