`2(x^2)e^(2x) + 2xe^(2x) = 0` Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility.

Textbook Question

Chapter 3, 3.4 - Problem 73 - Precalculus (3rd Edition, Ron Larson).
See all solutions for this textbook.

1 Answer | Add Yours

kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

Given

`2(x^2)e^(2x) + 2xe^(2x) = 0`

=> `(e^(2x)) *(2(x^2) +2x)=0`

=>`2x(e^(2x)) *(x +1) =0`

=>` 2x(e^(2x)) = 0 or (x+1) =0`

=> as `(e^(2x))` cannot be zero so, `x = 0 `

and in `(x+1) =0 => x = -1`

so the values of x are `x = 0 , x= -1`

the graphical representation of

`2(x^2)e^(2x) + 2xe^(2x) = 0`

 and

`y= 2(x^2)e^(2x) + 2xe^(2x)` are as follows

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 2)
This image has been Flagged as inappropriate Click to unflag
Image (2 of 2)

We’ve answered 318,917 questions. We can answer yours, too.

Ask a question