Question

How do you find the roots of `x^3-18x+27=0`?

Answer 1

The roots are `(3, -3/2+3sqrt5/2, -3/2-3sqrt5/2)`

Explanation:

Let `f(x)=x^3-18x+27`
Then by trial and error, `f(3)=27-54+27=0`
so `x=3` is a root
Then we do a long division
`x^3` `color(white)(aaaaa)` `-18x+27` `∣` `x-3`
`x^3-3x^2` `color(white)(aaaaaaaaa)` `∣` `x^2+3x-9`
`0+3x^2-18x`
`color(white)(aaaa)` `0-9x`
`color(white)(aaaaaa)` `-9x+27`
`color(white)(aaaaaa)` `-9x+27`
`color(white)(aaaaaaaa)` `0+0`
To find the other roots, we must solve
`x^x+3x-9=x^2+3x+9/4-9-9/4`
`(x+3/2)^2-45/4`
So `(x+3/2)^2=45/4` `=>` `x+3/2=+-sqrt45/2`
`x=-3/2+-3sqrt5/2`