a2x + a4 = ax+1 + ax+3
To solve this by inspection, note that when x=1 and x=3 , the LHS =RHS
So the solutions are x=1 and x=3 for all values of a?
My problem is how do we prove this more formally and can we be convinced that there are only two solutions?
And is inspection ok to do here or would you recommend against it?
"Inspection" here means essentially "trial and error, using obvious guesses". That is a fine way to find
some solutions. My next step might be to graph the two sides for a particular value of a and get a feel for how the equation works, which strongly suggests that for any given value of a, there will be only two solutions (which are, necessarily, the two that you found already that work for any a).
To prove this, I would probably use calculus, hoping to show that the derivative of the difference, a
2x + a
4 - a
x+1 - a
x+3, is such that there can be only two zeros. But let's see if we can do more.
One interesting feature of the equation is that 2x + 4 = (x + 1) + (x + 3). Another way to put it is that the average of the exponents on each side is the same. That suggests doing a substitution, letting u = x + 2, which is that average. Then x = u - 2, so the equation transforms to
a
2u-4 + a
4 = a
u-1 + a
u+1
Dividing everything by a
u, we get
a
u-4 + a
4-u = a
-1 + a.
At this point, I see that it can actually be solved algebraically, and we will get two solutions! That in itself answers your question.
I've completed the work, but decided it was too much fun to just show it all to you (especially if this is something for which you could get some sort of credit). See what you can do taking it from here, and I can give more hints as needed. (I made a couple more changes of variables).
Where did the problem come from?
