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

##### 1 Answer

#### Answer:

See explanation...

#### Explanation:

Given:

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

We can make this into a cubic polynomial using the substitution

#at^3+rt^2+s = 0#

Then we can solve this cubic to give three values of

One thing to watch out for is that because of the conventions for the principal square root of

It is possible for a cubic of the form

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

In practice, if

#st^3+rt+a = 0#

This can then be solved using Cardano's method or by using a trigonometric (