How do you solve x^2-2/3x-26/9x2−23x−269 by completing the square?

1 Answer
Shwetank Mauria
Dec 2, 2016

x^2-2/3x-26/9=(x-1/3+sqrt3)(x-1/3-sqrt3)x2−23x−269=(x−13+√3)(x−13−√3)

Explanation:

This is a polynomial and not an equation and hence cannot be solved for xx. However, we can factorize it using completing the square, as follows.

x^2-2/3x-26/9x2−23x−269

= ul(x^2-2xx(1/3)xx x+(1/3)^2)-(1/3)^2-26/9

= (x-1/3)^2-1/9-26/9

= (x-1/3)^2-27/9

= (x-1/3)^2-3

= (x-1/3)^2-(sqrt3)^2

and as a^2-b^2=(a+b)(a-b) hence above is equal to

= (x-1/3+sqrt3)(x-1/3-sqrt3)

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