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

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

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Answer 1 · 2017-07-01T23:24:16

$x = - 1 \pm \sqrt{\frac{5}{3}}$ $x = -1 +-sqrt(5/3)$

Explanation:

We have:

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

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

Step 1 - Factor out the quadratic coefficient, thus:

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

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

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

So now returning to the quadratic equation , we have;

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

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

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

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