How do you solve the equation $3 x^{2} + 5 x + 4 = 0$ $3x^2+5x+4=0$ by completing the square?

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How do you solve the equation $3 x^{2} + 5 x + 4 = 0$ $3x^2+5x+4=0$ by completing the square?

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Answer 1 · 2016-12-21T00:29:53

$x = \frac{1}{6} (- 5 \pm \sqrt{23} i)$ $x = 1/6(-5+-sqrt(23)i)$

Explanation:

Given:

$f (x) = 3 x^{2} + 5 x + 4$ $f(x) = 3x^2+5x+4$

I would first premultiply by $3 \cdot 2^{2} = 12$ $3*2^2 = 12$ to avoid having to deal with fractions very much...

$0 = 12 f (x)$ $0 = 12f(x)$

$0 = 12 (3 x^{2} + 5 x + 4)$ $color(white)(0) = 12(3x^2+5x+4)$

$0 = 36 x^{2} + 60 x + 48$ $color(white)(0) = 36x^2+60x+48$

$0 = 36 x^{2} + 60 x + 25 + 23$ $color(white)(0) = 36x^2+60x+25+23$

$0 = {(6 x + 5)}^{2} + 23$ $color(white)(0) = (6x+5)^2+23$

$0 = {(6 x + 5)}^{2} - {(\sqrt{23} i)}^{2}$ $color(white)(0) = (6x+5)^2-(sqrt(23)i)^2$

$0 = ((6 x + 5) - \sqrt{23} i) ((6 x + 5) + \sqrt{23} i)$ $color(white)(0) = ((6x+5)-sqrt(23)i)((6x+5)+sqrt(23)i)$

$0 = (6 x + 5 - \sqrt{23} i) (6 x + 5 + \sqrt{23} i)$ $color(white)(0) = (6x+5-sqrt(23)i)(6x+5+sqrt(23)i)$

Hence:

$x = \frac{1}{6} (- 5 \pm \sqrt{23} i)$ $x = 1/6(-5+-sqrt(23)i)$

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How do you solve the equation $3 x^{2} + 5 x + 4 = 0$ $3x^2+5x+4=0$ by completing the square?

1 Answer

Explanation:

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How do you solve the equation $3 x^{2} + 5 x + 4 = 0$ $3x^2+5x+4=0$ by completing the square?