# `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.

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### 1 Answer

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

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