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Chapter 3: Problem 74

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$-x^{2} e^{-x}+2 x e^{-x}=0$$

Short Answer

Expert verified
The solutions to the equation are \(x=0\) and \(x=2\).

Step by step solution

01

Factoring

Begin by factoring - it is noticeable that both terms contain common elements: Specifically -x and \(e^{-x}\). Factoring this out we get \(x e^{-x}(-x+2) = 0\).
02

Solving the Factors

Set each factor equal to zero, this gives us the equations: \(x=0\) and \(-x+2=0\). Solving the latter for x gives \(x=2\).
03

Checking the Answer graphically

To confirm these solutions are correct, plot the function \(-x^{2} e^{-x}+2 x e^{-x}\) and check that it crosses the x-axis (where y = 0) at x=0 and x=2.
04

Rounding the Final Result

Seeing as there are no decimals in the derived solutions, there is no need to round down to three decimal places. The result remains at \(x=0\) and \(x=2\).

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