document.write( "Question 241405: I am trying to solve (2x^2)*(e^2x)-(5xe^2x)=3e^(2x)\r
\n" ); document.write( "\n" ); document.write( "this has completely stumped me. \n" ); document.write( "

Algebra.Com's Answer #176788 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"2x%5E2e%5E%282x%29-5xe%5E%282x%29=3e%5E%282x%29\"$
\n" ); document.write( "One way to solve any complicated equation like this is to get one side equal to zero and factoring it. (This is one the of the techniques you learn when learning to solve quadratic equations.) So we'll start by subtracting $\"3e%5E%282x%29\"$ from each side:
\n" ); document.write( " $\"2x%5E2e%5E%282x%29-5xe%5E%282x%29-3e%5E%282x%29=0\"$
\n" ); document.write( "Now we'll factor. As usual, always start factoring by factoring out the GCF (unless it is 1). The GCF here is $\"e%5E%282x%29\"$ :
\n" ); document.write( " $\"e%5E%282x%29%282x%5E2-5x-3%29=0\"$
\n" ); document.write( "The second factor is a trinomial that will factor, too:
\n" ); document.write( " $\"e%5E%282x%29%282x%2B1%29%28x-3%29=0\"$
\n" ); document.write( "Now, according the the Zero Product Property, this (or any) product can be zero only if one (or more) of the factors is zero. So:
\n" ); document.write( " $\"e%5E%282x%29+=+0\"$ or $\"2x%2B1+=+0\"$ or $\"x-3+=+0\"$
\n" ); document.write( "Next we solve each of these. $\"e%5E%282x%29\"$ can never be zero, no matter what x is. So there will no solutions from the first equation. The other two equations, however, do have solutions:
\n" ); document.write( " $\"x+=+-1%2F2\"$ or $\"x+=+3\"$
\n" ); document.write( " \n" ); document.write( "

\n" );