How do you solve #x^2-2/3x-26/9# by completing the square?

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Answer 1 · 2016-12-02T15:07:35

#x^2-2/3x-26/9=(x-1/3+sqrt3)(x-1/3-sqrt3)#

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

#x^2-2/3x-26/9#

= #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)#