How do you solve #e^(2x)+9e^x+36=0#?

1 Answer
Feb 12, 2017

No Real solution exists.
Complex solutions do exist (but I don't think that's covered in the PreCalc courses)

Explanation:

Given that the equation is
#e^(2x)+9e^x+36=0#

Now at a quick glance, you'd think this is a quadratic equation and try to solve it as thus. But before you do so, take a step back and look at how the equation is. It's not just any ordinary quadratic equation, based on #x# and all, but instead on a power function.

Now, notice that all the values are added up and the equation says that they add up to zero. Let's check out all the values that are in the equation.

#36#, of course, is a positive number, this makes us conclude that to give us an equation summing up to zero, the other terms must be negative for some value of #x#.

But, the other terms are power function, and for any value of #x#, #a^x>0# (#AA a>0#, which #e# is a part of). This means that for no real value of #x# do we get a negative value to make the equation zero.

If OP wanted the complex answers, do ask in the comments.