need help verifying answers
find the zeros of : f(x)=x^2 (2e^2x) +(2xe^2x) + (e^2x) + (2xe^2x)

I get : x= -1+ (1/2)sqrt 2 and x=-1-(1/2)sqrt 2

Question

need help verifying answers 
find the zeros of : f(x)=x^2 (2e^2x) +(2xe^2x) + (e^2x) + (2xe^2x)

I get :   x= -1+ (1/2)sqrt 2  and x=-1-(1/2)sqrt 2

blurbendy · Answer

x = 0

x = -1 -1/Sqrt[2]

x = 1/Sqrt[2] - 1

anonymous · Answer

i dont distribute he x^2  therefore x=0 and the other two change as well?

blurbendy · Answer

wait, what?

anonymous · Answer

$$f(x)= x^2 (2e^{2x})+2xe ^{2x}+e ^{2x}+2xe ^{2x}$$

blurbendy · Answer

Yes, i know.  trust me, my answer is correct.  plug the values in if you don't believe me.

anonymous · Answer

yea i trust you, my question is you set all of that equal to 0.  i distributed the x^2  did you?

anonymous · Answer

take e^2x common 

$$\Large f(x)=2e^{2x}(x^2+2x+1/2)$$
set it equal to zero
$$\Large 2e^{2x}=0$$    no solution can be obtaind for this .
so solve 
$$\Large  x^2+2x+1/2=0$$

by completing square
x^2+2x =-1/2
x^2+2x+1=1/2
(x+1)^2=1/2
take square root of both sides 

$$\Large  x=-1-\frac{1}{\sqrt{2}}$$

$$\Large  x=\frac{1}{\sqrt{2}}-1$$

these are the  possible  zeros .

anonymous · Answer

in the first step  take  2e^2x common !