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

2 Answers
Jun 20, 2018

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))

Jun 20, 2018

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.