Question

Can the quadratic formula be used with functions of the form `f(x)=ae^(bx)+ce^((b/2)x)+d`? Why or why not?

Answer 1

Yes - notice that it is a quadratic in the exponential function

Explanation:

`f(x)=ae^(bx)+ce^((b/2)x)+d=a(e^((b/2)x))^2+c(e^((b/2)x))+d`

The equation is a quadratic in `e^((b/2)x)`, so (by the quadratic formula)

`e^((b/2)x)=(-c+-sqrt(c^2-4ad))/(2a)`

and hence

`x=2/bln((-c+-sqrt(c^2-4ad))/(2a))`

Answer 2

Yes

Explanation:

If you let `e^{b/2x} = t`, then you have

`t^2 = (e^{b/2x})^2 = e^{2*b/2x} = e^(bx)`

So, the equation becomes

`at^2+ct+d`, which you can indeed solve with the quadratic equation.

Then, assuming you found two solutions `t_1` and `t_2`, you have to go back to `x` values: you have

`t = t_1 \iff e^{b/2x} = t_1 \iff b/2x = log(t_1) \iff x = 2log(t_1)/b`

and similarly for `t_2`.

This means that you might find solutions in `t` that are no compatible with `x`. For example, if `t_1=-5`, you would have

`x = 2/b log(-5)`

which is not possible to compute using real numbers.