What are all the real zeros of #y=(x-12)^3-10#?

Question

4481 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-20T12:42:24

The real zero is:

#x = 12+root(3)(10)#

Given:

#y = (x-12)^3-10#

The real zero is:

#x = 12+root(3)(10)#

The other zeros are both non-real complex:

#x = 12+omega root(3)(10)#

#x = 12+omega^2 root(3)(10)#

where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.