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

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Answer 1 · 2016-10-24T05:23:16

$x=2/3+sqrt(10)/3$ , $x=2/3-sqrt(10)/3$

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

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

$(3x^2-4x-2)/3=0/3$

$x^2-4/3x-2/3=0$

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

$x^2+ax=(x+a/2)^2-(a/2)^2$

so using this formula,

$x^2-4/3x-2/3=0$

goes to

$(x-4/6)^2-16/36-2/3=0$

simplifying down,

$(x-2/3)^2-10/9=0$

rearrange for x,

$(x-2/3)^2= 10/9$

$x-2/3= +-sqrt(10/9)$

$x=2/3+sqrt(10)/3$ , $x=2/3-sqrt(10)/3$

these are the two solutions.

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