Question

Can someone please explain How do I solve this 2xe^x+x^2e^x= 0

e^3x+3xe^3x=0

2x^2 e^x^2- e^x^2 / x^2 =0

Answer 1

just put all the numbers together

Answer 2

33322222 that would be my answer

Answer 3

what do have to do?

Answer 4

put all the numbers you see together

Answer 5

Solve the equations. There is three seperate equations so you can't put all the numbers together

Answer 6

oh if it was me i would just put what i see

Answer 7

\[2xe ^{x} +x^2 e^x =0\]

Answer 8

so 0

Answer 9

no. What ?

Answer 10

for the 1st question, take out the common factor of e^x

then solve the quadratic

\[e^{3x}(1 + 3x) = 0\]

the problem is that \[e^{3x} \neq 0\]

as you can't find a value of x that make e^(3x) zero.

so then by taking the common factor you can find the value of x that makes

1 + x = 0

because if you substitute that value into the equation you would have

e^{3x}*(0) = 0

hope this helps

Answer 11

Thank you it does

Answer 12

is the 2nd question

1.

\[2x^2e^{x^2} - \frac{e^{x^2}}{x^2} = 0\]

or

2.

\[\frac{2x^2 e^{x^2} - e^{x^2}}{x^2}=0\]

Answer 13

and for the 1st question take e^x as a common factor and you have

\[e^x(2x + x^2) = 0\]

solve the quadratic.... as the exponential will never equal zero

Answer 14

The whole thing is being divided by x^2

Answer 15

ok.. so factor the numerator

you get

\[\frac{e^{x^2}(2x^2 -1)}{x^2}\]

so now looking at the problem, the numerator can't be zero

so just solve the quadratic

\[2x^2 - 1 = 0\]

Answer 16

Okay thank you so much