How do you solve n^2-4/3n-14/9=0n2−43n−149=0 by completing the square?

1 Answer
Shwetank Mauria
Jun 8, 2017

n=2/3-sqrt2n=23−√2 or n=2/3-sqrt2n=23−√2

Explanation:

n^2-4/3n-14/9=0n2−43n−149=0

i.e. n^2-2xx2/3xxn-14/9=0n2−2×23×n−149=0

Now, as (a-b)^2=a^2-2xxaxxb+b^2(a−b)2=a2−2×a×b+b2, comparing it with n^2-2xx2/3xxnn2−2×23×n, we find adding (2/3)^2=4/9(23)2=49, will complete the square and we have

n^2-2xx2/3xxn+(2/3)^2-4/9-14/9=0n2−2×23×n+(23)2−49−149=0

or (n-2/3)^2-18/9=0(n−23)2−189=0

or (n-2/3)^2-2=0(n−23)2−2=0

or (n-2/3)^2-(sqrt2)^2=0(n−23)2−(√2)2=0 and using a^2-b^2=(a+b)(a-b)a2−b2=(a+b)(a−b)

it becomes (n-2/3+sqrt2) (n-2/3-sqrt2)=0(n−23+√2)(n−23−√2)=0

Hence either n-2/3+sqrt2=0n−23+√2=0 i.e. n=2/3-sqrt2n=23−√2

or n-2/3-sqrt2=0n−23−√2=0 i.e. n=2/3+sqrt2n=23+√2

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