Can the quadratic formula be used with functions of the form f(x)=ae^(bx)+ce^((b/2)x)+df(x)=aebx+ce(b2)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))+df(x)=aebx+ce(b2)x+d=a(e(b2)x)2+c(e(b2)x)+d

The equation is a quadratic in e^((b/2)x)e(b2)x, so (by the quadratic formula)

e^((b/2)x)=(-c+-sqrt(c^2-4ad))/(2a)e(b2)x=c±c24ad2a

and hence

x=2/bln((-c+-sqrt(c^2-4ad))/(2a))x=2bln(c±c24ad2a)

Jun 20, 2018

Yes

Explanation:

If you let e^{b/2x} = teb2x=t, then you have

t^2 = (e^{b/2x})^2 = e^{2*b/2x} = e^(bx)t2=(eb2x)2=e2b2x=ebx

So, the equation becomes

at^2+ct+dat2+ct+d, which you can indeed solve with the quadratic equation.

Then, assuming you found two solutions t_1t1 and t_2t2, you have to go back to xx values: you have

t = t_1 \iff e^{b/2x} = t_1 \iff b/2x = log(t_1) \iff x = 2log(t_1)/bt=t1eb2x=t1b2x=log(t1)x=2log(t1)b

and similarly for t_2t2.

This means that you might find solutions in tt that are no compatible with xx. For example, if t_1=-5t1=5, you would have

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

which is not possible to compute using real numbers.