# How do you solve equations of the form ax^(3/2)+rx+s = 0 ?

Jan 5, 2018

See explanation...

#### Explanation:

Given:

$a {x}^{\frac{3}{2}} + r x + s = 0$

We can make this into a cubic polynomial using the substitution $t = {x}^{\frac{1}{2}}$ to get:

$a {t}^{3} + r {t}^{2} + s = 0$

Then we can solve this cubic to give three values of $t$ and hence of $x = {t}^{2}$.

One thing to watch out for is that because of the conventions for the principal square root of $x$, the only values for $t$ that yield solutions of the original equation are complex (possibly real) numbers with $A r g \left(t\right) \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

It is possible for a cubic of the form $a {t}^{3} + r {t}^{2} + s$ to have $0 , 1$ or $2$ roots in the required range, yielding $0 , 1$ or $2$ complex (possibly real) solutions to the original equation. Note that $3$ suitable roots is not possible, since the sum of the reciprocals of the roots must be $0$.

In particular, if the cubic has two positive real solutions and one negative one, then the original equation has exactly two real solutions and no others. For example:

$0 = \left(t - 1\right) \left(2 t - 1\right) \left(3 t + 1\right)$

$\textcolor{w h i t e}{0} = \left(2 {t}^{2} - 3 t + 1\right) \left(3 t + 1\right)$

$\textcolor{w h i t e}{0} = 6 {t}^{3} - 7 {t}^{2} + 1$

So:

$6 {x}^{\frac{3}{2}} - 7 x + 1 = 0$

has exactly two solutions, both of which are real, namely $x = 1$ and $x = \frac{1}{4}$.

In practice, if $s \ne 0$, then to get a simpler cubic to solve, use the substitution $t = {x}^{- \frac{1}{2}}$ and solve this cubic instead:

$s {t}^{3} + r t + a = 0$

This can then be solved using Cardano's method or by using a trigonometric ($t = k \cos \theta$) substitution.