How do you solve using completing the square method $3x^2 + 5x = -x + 4$ ?

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Answer 1 · 2018-06-26T10:23:07

$x = -1 +- sqrt(7/3)$

$3x^2 + 5x = - x + 4$

$3x^2 + 6x = 4$

$(sqrt3 x)^2 + 2 * sqrt3 * sqrt3 x + (sqrt3)^2 = 4 + (sqrt3)^2$

$(sqrt3 x + sqrt3)^2 = 7$

$(sqrt3 x + sqrt3) = +-(sqrt7)$

$x + 1 = sqrt(7/3)$

$x = -1 +- sqrt(7/3)$

Answer 2 · 2018-06-26T10:26:06

$x = -1 +- sqrt(7/3)$

$3x^2 + 5x = -x + 4$

$3x^2 + 6x = 4$

$x^2 + 2x = 4/3$

$x^2 + 2x + 1 = 4/3 = 1 = 7/3$

$(x + 1)^2 = (sqrt(7/3))^2$

$x = 1 = +- sqrt(7/3)$

$x = -1 +- sqrt(7/3)$

How do you solve using completing the square method 3x^2 + 5x = -x + 43x^2 + 5x = -x + 4?