How do you complete the square $3 x^{2} - 6 x - 1 = 0$ $3x^2-6x-1=0$ ?

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Answer 1 · 2015-02-04T18:22:31

The answer is:

$3 (x^{2} - 2 x) - 1 = 0 \Rightarrow 3 (x^{2} - 2 x + 1 - 1) - 1 = 0 \Rightarrow$ $3(x^2-2x)-1=0rArr3(x^2-2x+1-1)-1=0rArr$

$3 (x^{2} - 2 x + 1) - 3 - 1 = 0 \Rightarrow 3 {(x - 1)}^{2} - 4 = 0 \Rightarrow$ $3(x^2-2x+1)-3-1=0rArr3(x-1)^2-4=0rArr$

${(x - 1)}^{2} = \frac{4}{3} \Rightarrow x - 1 = \pm \sqrt{\frac{4}{3}} = 0 \Rightarrow$ $(x-1)^2=4/3rArrx-1=+-sqrt(4/3)=0rArr$

$x = 1 \pm \frac{2}{\sqrt{3}} \Rightarrow x = 1 \pm \frac{2 \sqrt{3}}{3} \Rightarrow$ $x=1+-2/sqrt3rArrx=1+-(2sqrt3)/3rArr$

$x = \frac{3 \pm 2 \sqrt{3}}{3}$ $x=(3+-2sqrt3)/3$ .

How do you complete the square 3x2−6x−1=03x^2-6x-1=0?