Analytical solution for equation with power $n \in [1, 2]$

Question

Does there exist analytical solution to the equation $ax^p + bx + c = 0$ where $p \in [1, 2]$? Please provide references if the answer is affirmative.

Answer 1

Nope. Only numerical or approximate answers.

If you know something about the coefficients, you can possibly establish a region where the root(s) might be.

For example, if you can show that $f(x) = ax^p + bx + c$ satisfies $f'(x) > 0$, then it can have at most one root.

Then looking at $f(0)$ and $f(1)$ might provide a region where the root has to be.

Answer 2

After marty cohen's answer.

SInce we look for the zero, assuming $a \neq 0$, divide all terms by $a$ and consider that you look for the zero(s) of function $$f(x) = x^p + \beta x + \gamma$$ Because of the non integer value of $p$, we are restricted to the range $x> 0$.

The derivatives $$f'(x)=p\, x^{p-1}+\beta \qquad \text{and} \qquad f''(x)= (p-1) p \,x^{p-2}$$ The first derivative cancels at $$x_*=\left(-\frac{\beta}{ p}\right)^{\frac{1}{p-1}}$$ which is real if $\beta<0$. Since $f(0)=\gamma$, we also need $\gamma <0$.

If this is the case, there is one root. To approximate it, we can have an estimate building a second order Taylor expansion around $x=x_*$ to get $$x_0=x_* + \sqrt{-2 \frac {f(x_*)}{f''(x_*)}}$$

For illustation puproses, let us try $p=1.234$, $\beta=2.345$; $\gamma=-3456$. This will give $x_*=15.5443$ and $x_0=458.481$. Now, using Newton method, the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 458.481388028 \\ 1 & 1379.34417348 \\ 2 & 1196.22489568 \\ 3 & 1191.45000524 \\ 4 & 1191.44650168 \end{array} \right)$$

Notice that we have one overshoot of the solution because $f(x_0)\times f''(x_0) <0$ (Darboux theorem).

Answer 3

There are series solutions if $b/(cp)$ is small. If $w = (-c/a)^{1/p}$ and $v = b/(cp)$,

$$x = w + w^2 v + \frac{(3-p)}{2!} w^3 v^2 + \frac{(4-p)(4-2p)}{3!} w^4 v^3 + \frac{(5-p)(5-2p)(5-3p)}{4!} w^5 b^4 + \ldots $$