Question #fa7ed

1 Answer
Nityananda
Nov 12, 2016

[1-sqrt2]/3 and [1+sqrt2]/3

Explanation:

9x^2-6x+1 = 2
or, 9x^2-6x+1-2=0
or, (3x)^2-2.3x.1+(1)^2- [sqrt2]^2=0
or,(3x-1)^2-[sqrt2]^2=0
or, [3x-1+sqrt2][3x-1-sqrt2]=0
or, [3x-1+sqrt2] = 0, [3x-1-sqrt2]=0
or, 3x= 1-sqrt2, 3x= 1+sqrt2
or, x = [1-sqrt2]/3 and [1+sqrt2]/3

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