How do you solve using the completing the square method #3x^2-6x-1=0#?

1 Answer
Dec 26, 2016

Answer:

#x=2.155" or "x = -0.155#

Explanation:

Quadratic equations are in the form #ax^2 +bx+c =0#

#color(lime)(3)x^2 -6x-1=0" "larrdiv 3# to make #color(lime)(a=1)#
#color(lime)(darr)#
#color(lime)(1)x^2 -2x-1/3 =0" "larr# move #-1/3# to the RHS

#x^2 -2x" "=1/3#

#x^2 -2x color(blue)(+ square) = +1/3" "larrcolor(blue)( square= (b/2)^2)#
#color(white)(.......)color(blue)(darr)#
#" "color(blue)(((-2)/2)^2 = 1= square)" "larr# add to both sides

#x^2 -2x color(blue)(+1) = +1/3color(blue)(+1)" "larr#this is completing the square

[Now use the identity #color(red)(a^2 + 2ab+b^2 = (a+b)^2) # ]

#color(red)(x^2 -2x +1) = +1 1/3#
#color(white)(....)color(red)(darr)#
#color(red)((x-1)^2) = 1 1/3 =4/3" "#find square root both sides

#x-1 = +-sqrt(4/3)" "larr# solve for #x# twice

#x = +sqrt(4/3) +1" and "x = -sqrt(4/3)+1#

#x =2/sqrt3 +1" and "x = (-2)/sqrt3 +1#

#x=2.155" or "x = -0.155#