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

1 Answer
Mar 7, 2018

Answer:

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#