Question #fa7ed

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Question #fa7ed

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Answer 1 · 2016-11-12T17:36:07

$\frac{1 - \sqrt{2}}{3} and \frac{1 + \sqrt{2}}{3}$ $[1-sqrt2]/3 and [1+sqrt2]/3$

Explanation:

$9 x^{2} - 6 x + 1 = 2$ $9x^2-6x+1 = 2$
or, $9 x^{2} - 6 x + 1 - 2 = 0$ $9x^2-6x+1-2=0$
or, ${(3 x)}^{2} - 2.3 x .1 + {(1)}^{2} - {[\sqrt{2}]}^{2} = 0$ $(3x)^2-2.3x.1+(1)^2- [sqrt2]^2=0$
or, ${(3 x - 1)}^{2} - {[\sqrt{2}]}^{2} = 0$ $(3x-1)^2-[sqrt2]^2=0$
or, $[3 x - 1 + \sqrt{2}] [3 x - 1 - \sqrt{2}] = 0$ $[3x-1+sqrt2][3x-1-sqrt2]=0$
or, $[3 x - 1 + \sqrt{2}] = 0, [3 x - 1 - \sqrt{2}] = 0$ $[3x-1+sqrt2] = 0, [3x-1-sqrt2]=0$
or, $3 x = 1 - \sqrt{2}, 3 x = 1 + \sqrt{2}$ $3x= 1-sqrt2, 3x= 1+sqrt2$
or, $x = \frac{1 - \sqrt{2}}{3} and \frac{1 + \sqrt{2}}{3}$ $x = [1-sqrt2]/3 and [1+sqrt2]/3$

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Question #fa7ed

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