How do you solve $3x^2+6x-2=0$ by completing the square?

Question

19698 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-01T23:24:16

$x = -1 +-sqrt(5/3)$

We have:

$3x^2+6x-2=0$

The standard steps to complete the square on quadratic expression are as follows:

Step 1 - Factor out the quadratic coefficient, thus:

$3x^2+6x-2=3{x^2+2x-2/3}$

Step 2 - Factor $1/2$ of the $x$ coefficient to form a perfect square, and subtract its square, and simplifiy, thus:

$3x^2+6x-2 = 3{(x+2/2)^2-(2/2)^2-2/3}$
$" " = 3{(x+1)^2-1-2/3}$
$" " = 3{(x+1)^2-5/3}$

So now returning to the quadratic equation , we have;

$3x^2+6x-2=0$

$:. 3{(x+1)^2-5/3} = 0$
$:. (x+1)^2-5/3 = 0$
$:. (x+1)^2 = 5/3$
$:. x+1 = +-sqrt(5/3)$
$:. x = -1 +-sqrt(5/3)$

How do you solve 3x^2+6x-2=03x^2+6x-2=0 by completing the square?