How do you solve using the completing the square method #x^2 + 3x – 6 = 0#?

2 Answers
Jun 26, 2018

Answer:

#x=-3/2+-sqrt33/2#

Explanation:

#x^2+3x=6#

#x^2+2(3/2)xcolor(red)(+9/4)=6color(red)(+9/4)#

#(x+3/2)^2=33/4#

#color(blue)"take the square root of both sides"#

#sqrt((x+3/2)^2)=+-sqrt(33/4)larrcolor(blue)"note plus or minus"#

#x+3/2=+-sqrt33/2#

#"subtract "3/2" from both sides"#

#x=-3/2+-sqrt33/2larrcolor(red)"exact solutions"#

Jun 26, 2018

Answer:

#color(brown)(x = (-1/2) (3 - sqrt33 ), -(1/2) ( 3 + sqrt33)#

Explanation:

#x^2 + 3x - 6 = 0#

#x^2 + 3x + (3/2)^2 - 6 = (3/2)^2, "adding " (3/2)^2 " to both sides"#

#x^2 + 2 * 3/2 * x + (3/2)^2 = (3/2)^2 + 6#

#(x + 3/2)^2 = 33/4#

#(x + 3/2)^2 = (sqrt(33/4))^2#

#x + 3/2 = +- sqrt(33/4), " taking squre root on both sides"#

#x = -3/2 +- sqrt33 / 2#

#color(brown)(x = (-1/2) (3 - sqrt33 ), -(1/2) ( 3 + sqrt33)#