How do you solve $3 x^{2} - 4 x - 2 = 0$ $3x^2-4x-2=0$ by completing the square?

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Answer 1 · 2016-10-24T05:23:16

$x = \frac{2}{3} + \frac{\sqrt{10}}{3}$ $x=2/3+sqrt(10)/3$ , $x = \frac{2}{3} - \frac{\sqrt{10}}{3}$ $x=2/3-sqrt(10)/3$

Firstly, we need to deal with the coefficient of $x^{2}$ $x^2$ (the 3).

In order to remove the 3 we divide the whole expression by 3.

$\frac{3 x^{2} - 4 x - 2}{3} = \frac{0}{3}$ $(3x^2-4x-2)/3=0/3$

$x^{2} - \frac{4}{3} x - \frac{2}{3} = 0$ $x^2-4/3x-2/3=0$

next, we complete the square, here is the general formula

$x^{2} + a x = {(x + \frac{a}{2})}^{2} - {(\frac{a}{2})}^{2}$ $x^2+ax=(x+a/2)^2-(a/2)^2$

so using this formula,

$x^{2} - \frac{4}{3} x - \frac{2}{3} = 0$ $x^2-4/3x-2/3=0$

goes to

${(x - \frac{4}{6})}^{2} - \frac{16}{36} - \frac{2}{3} = 0$ $(x-4/6)^2-16/36-2/3=0$

simplifying down,

${(x - \frac{2}{3})}^{2} - \frac{10}{9} = 0$ $(x-2/3)^2-10/9=0$

rearrange for x,

${(x - \frac{2}{3})}^{2} = \frac{10}{9}$ $(x-2/3)^2= 10/9$

$x - \frac{2}{3} = \pm \sqrt{\frac{10}{9}}$ $x-2/3= +-sqrt(10/9)$

$x = \frac{2}{3} + \frac{\sqrt{10}}{3}$ $x=2/3+sqrt(10)/3$ , $x = \frac{2}{3} - \frac{\sqrt{10}}{3}$ $x=2/3-sqrt(10)/3$

these are the two solutions.

How do you solve 3x^2-4x-2=03x2−4x−2=03x^2-4x-2=0 by completing the square?