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

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Answer 1 · 2018-03-03T09:00:35

#color(blue)(x = 1/3, -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#