Chapter 3.4, Problem 74E

To determine

To find : solution to the equation algebraically and verify the result graphically.

Answer to Problem 74E

The solution to the equation $-x^2{e}^{-x}+2x{e}^{-x}=0$ are $\underset{\_}{\boxed{x=0\&2}}$

Explanation of Solution

Given information: $-x^2{e}^{-x}+2x{e}^{-x}=0$

Concept Involved:

The word “solve” means the process of finding the value of x that makes the equation true. In order to solve we need to undo whatever is done to x . The solution to the equation $f\left(x\right)=g\left(x\right)$ graphically is the x value of the point of intersection of where the two graphs meet.

Formula Used:

  $\ln m^n=n\ln m$

Zero factor property: If $a\cdot b=0$ , then $a=0\left(or\right)b=0$

Calculation:

Factor the Greatest Common Factor in the left side of the equation.

  $-x\cdot e^{-x}\left(x-2\right)=0$

Use the zero factor property to set each factor to zero

  $-x=0$▶ 1^st equation

  $e^{-x}=0$ ▶ 2^nd equation

  $x-2=0$ ▶ 3^rd equation

Solving the equation $-x=0$

• By dividing -1 on both sides, and then
• By Simplifying fraction on both sides
$\frac{-x}{-1}=\frac{0}{-1}$

  $x=0$

Solving the equation $e^{-x}=0$

• There is no value of x that makes the above equation TRUE, so No solution

Solving the equation $x-2=0$

• By adding 2 on both sides, and then
• By Simplifying on both sides
$x-2+2=0+2$

  $x=2$

Check for extraneous solution by substituting the result in the original equation.

  $x=0$ makes the original equation TRUE.

  $x=2$ makes the original equation TRUE.

Graph:

  

Interpretation:

Setting up left side of the equation $f\left(x\right)=-x^2{e}^{-x}+2x{e}^{-x}$ and right side of the equation $g\left(x\right)=0$ .

The xcoordinate of point of intersection of two graphs is the solution to the equation $-x^2{e}^{-x}+2x{e}^{-x}=0$

Conclusion:

The solution to the equation $-x^2{e}^{-x}+2x{e}^{-x}=0$ are $\underset{\_}{\boxed{x=0\&2}}$