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

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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. $n=2/3+sqrt2$

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