SOLUTION: Find the zeros of f. (Enter your answers as a comma-separated list.) f(x) = x^2(2e^6x) + 4xe^6x + 7e^6x + 4xe^6x I cancelled out the e^6x and set the equation equal to zero and

Algebra ->  Equations -> SOLUTION: Find the zeros of f. (Enter your answers as a comma-separated list.) f(x) = x^2(2e^6x) + 4xe^6x + 7e^6x + 4xe^6x I cancelled out the e^6x and set the equation equal to zero and      Log On


   

Question 1156555: Find the zeros of f. (Enter your answers as a comma-separated list.)
f(x) = x^2(2e^6x) + 4xe^6x + 7e^6x + 4xe^6x
I cancelled out the e^6x and set the equation equal to zero and got 2x^2+8x+7 which I cannot factor so i used the quadratic formula and got -8+2(root 2)/4 and 8+2(root 2)/4 which is not correct either. HELP!
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

You DO NOT NEED factor the remaining equation, after canceling  e%5E%286x%29.


Solve it using the quadratic formula.


NOBODY asks you to factor the remained polynomial.


Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29+=+x%5E2%282e%5E%286x%29%29+%2B+4xe%5E%286x%29+%2B+7e%5E%286x%29+%2B+4xe%5E%286x%29
f%28x%29+=0
2x%5E2%2Ae%5E%286x%29+%2B+4xe%5E%286x%29+%2B+7e%5E%286x%29+%2B+4xe%5E%286x%29=0
%282x%5E2+%2B+4x+%2B+7+%2B+4x%29e%5E%286x%29=0
%282x%5E2+%2B+8x+%2B+7+%29e%5E%286x%29=0
e%5E%286x%29=0+=>no solutions exist
2x%5E2+%2B+8x+%2B+7+=0......use quadratic formula
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
x=%28-8%2B-sqrt%288%5E2-4%2A2%2A7%29%29%2F%282%2A2%29
x=%28-8%2B-sqrt%2864-56%29%29%2F4
x=%28-8%2B-sqrt%288%29%29%2F4
x=%28-8%2B-sqrt%284%2A2%29%29%2F4
x=%28-8%2B-2sqrt%282%29%29%2F4
x=%28-4%2B-sqrt%282%29%29%2F2
solutions:
x=%28-4%2Bsqrt%282%29%29%2F2
x=-2%2Bsqrt%282%29%2F2...rationalize:
highlight%28x+=+-2+%2B+1%2Fsqrt%282%29%29

x=%28-4-sqrt%282%29%29%2F2...rationalize, same as above
highlight%28x=-2-1%2Fsqrt%282%29%29