Question

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

Answer 1

`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`