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

Jun 20, 2018

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

#### Explanation:

$f \left(x\right) = a {e}^{b x} + c {e}^{\left(\frac{b}{2}\right) x} + d = a {\left({e}^{\left(\frac{b}{2}\right) x}\right)}^{2} + c \left({e}^{\left(\frac{b}{2}\right) x}\right) + d$

The equation is a quadratic in ${e}^{\left(\frac{b}{2}\right) x}$, so (by the quadratic formula)

${e}^{\left(\frac{b}{2}\right) x} = \frac{- c \pm \sqrt{{c}^{2} - 4 a d}}{2 a}$

and hence

$x = \frac{2}{b} \ln \left(\frac{- c \pm \sqrt{{c}^{2} - 4 a d}}{2 a}\right)$

Jun 20, 2018

Yes

#### Explanation:

If you let ${e}^{\frac{b}{2} x} = t$, then you have

${t}^{2} = {\left({e}^{\frac{b}{2} x}\right)}^{2} = {e}^{2 \cdot \frac{b}{2} x} = {e}^{b x}$

So, the equation becomes

$a {t}^{2} + c t + d$, 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

$t = {t}_{1} \setminus \iff {e}^{\frac{b}{2} x} = {t}_{1} \setminus \iff \frac{b}{2} x = \log \left({t}_{1}\right) \setminus \iff x = 2 \log \frac{{t}_{1}}{b}$

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

$x = \frac{2}{b} \log \left(- 5\right)$

which is not possible to compute using real numbers.