How do you solve #n^2-4/3n-14/9=0# by completing the square?

Question

1523 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-08T05:17:16

#n=2/3-sqrt2# or #n=2/3-sqrt2#

#n^2-4/3n-14/9=0#

i.e. #n^2-2xx2/3xxn-14/9=0#

Now, as #(a-b)^2=a^2-2xxaxxb+b^2#, comparing it with #n^2-2xx2/3xxn#, we find adding #(2/3)^2=4/9#, will complete the square and we have

#n^2-2xx2/3xxn+(2/3)^2-4/9-14/9=0#

or #(n-2/3)^2-18/9=0#

or #(n-2/3)^2-2=0#

or #(n-2/3)^2-(sqrt2)^2=0# and using #a^2-b^2=(a+b)(a-b)#

it becomes #(n-2/3+sqrt2) (n-2/3-sqrt2)=0#

Hence either #n-2/3+sqrt2=0# i.e. #n=2/3-sqrt2#

or #n-2/3-sqrt2=0# i.e. #n=2/3+sqrt2#