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

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Answer 1 · 2018-06-20T13:01:28

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

#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 · 2018-06-20T13:10:05

Yes

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.