Question

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

Answer 1

See explanation...

Explanation:

Given:

`ax^(3/2)+rx+s = 0`

We can make this into a cubic polynomial using the substitution `t=x^(1/2)` to get:

`at^3+rt^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 `Arg(t) in (-pi/2, pi/2]`.

It is possible for a cubic of the form `at^3+rt^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 = (t-1)(2t-1)(3t+1)`

`color(white)(0) = (2t^2-3t+1)(3t+1)`

`color(white)(0) = 6t^3-7t^2+1`

So:

`6x^(3/2)-7x+1 = 0`

has exactly two solutions, both of which are real, namely `x=1` and `x=1/4`.

In practice, if `s!=0`, then to get a simpler cubic to solve, use the substitution `t = x^(-1/2)` and solve this cubic instead:

`st^3+rt+a = 0`

This can then be solved using Cardano's method or by using a trigonometric (`t = kcos theta`) substitution.