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

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

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Answer 1 · 2017-09-10T04:52:08

$x = \pm \frac{i \sqrt{- 5}}{3} - \frac{2}{3}$ $x= +-(isqrt(-5))/3-2/3$

Explanation:

Given -

$3 x^{2} + 4 x + 3 = 0$ $3x^2+4x+3=0$
Take the constant term to right

$3 x^{2} + 4 x = - 3$ $3x^2+4x=-3$

Divide each term on both sides by the coefficient of $x^{2}$ $x^2$

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

$x^{2} + \frac{4}{3} x = - 1$ $x^2+4/3 x=-1$

Divide the coefficient $x$ $x$ , square it and add it to both sides

$x^{2} + \frac{2}{3} x + \frac{4}{9} = - 1 + \frac{4}{9} = \frac{- 9 + 4}{9} = - \frac{5}{9}$ $x^2+2/3 x+4/9=-1+4/9=(-9+4)/9=-5/9$

${(x + \frac{2}{3})}^{2} = - \frac{5}{9}$ $(x+2/3)^2=-5/9$
Take square root on both sides

$(x + \frac{2}{3}) = \sqrt{- \frac{5}{9}} = \frac{\sqrt{- 5}}{3}$ $(x+2/3)=sqrt(-5/9)=sqrt(-5)/3$

$x = \pm \frac{i \sqrt{- 5}}{3} - \frac{2}{3}$ $x= +-(isqrt(-5))/3-2/3$

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

1 Answer

Explanation:

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