How do you solve #(x - 1/3)^2 = 4/9#?

Question

1489 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-08T05:08:49

#x=1# or #x=-1/3#

#(x-1/3)^2=4/9#

root square

#abs(x-1/3)=sqrt(4/9)#

#abs(x-1/3)=2/3#

now if #abs(a)=b# then #a=b or a=-b#

#x-1/3=2/3# or #x-1/3=-2/3#

#x=1# or #x=-1/3#