How do you solve $3 x^{2} + 17 x - 6 = 0$ $3x^2 + 17x -6 = 0$ by completing the square?

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Answer 1 · 2018-03-03T09:00:35

$x = \frac{1}{3}, - 6$ $color(blue)(x = 1/3, -6$

$3 x 2 + 17 x - 6$ $3x2 + 17x - 6$ 0#

$cancel3(x^2 + (17/3)x -2) = 0$

$x^2 + color(purple)((2 * (17/6) x) ) = 2$ to make the middle term in the form of $2xy$

Add $color(green)( (17/6)^2$ to both sides to make the L H S a perfect square.

$x^2 + (2 * (17/6)x + (17/6)^2 = 2 + (17/6)^2 = 361/36$

$(x+ (17/6))^2 = (19/6)^2$

$x + 17/6 =color(red)( +-) 19/6$

$x =color(red)( +-) 19/6 - 17/6$

$color(blue)(x = 1/3, -6$

How do you solve 3x^2 + 17x -6 = 03x2+17x−6=03x^2 + 17x -6 = 0 by completing the square?