Question

How do you find the real zeros of `y=3/4(x-2/3)^3-16/9`?

Answer 1

2

Explanation:

You rewrite `3/4(x-2/3)^3-16/9 = 0` in the form :

`3/4(x-2/3)^3=16/9`

so that

`(x-2/3)^3=16/9 times 4/3 = (4/3)^3`

The only real solution to `x^3=a^3` is `x=a` (the other two are `x=a omega` and `x=a omega^2`, where `omega=exp({2pi}/3i)` and `omega^2` are the complex cube roots of unity), so

`x-2/3 = 4/3`
and thus
`x = 2/3+4/3=2`