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

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.