Question

How do you solve the equation `6x^2-17x+12=0` by completing the square?

Answer 1

`x = 3/2" "` and `" "x = 4/3`

Explanation:

To cut down on arithmetic involving fractions, multiply by `24 = 6*2^2` in order to make the leading term a perfect square and provide a factor `2` for the middle coefficient.

Once we reach a difference of squares, use the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with `A=(12x-17)` and `B=1`.

So:

`0 = 24(6x^2-17x+12)`

`color(white)(0) = 144x^2-408x+288`

`color(white)(0) = (12x)^2-2(12x)(17)+17^2-1`

`color(white)(0) = (12x-17)^2-1^2`

`color(white)(0) = ((12x-17)-1)((12x-17)+1)`

`color(white)(0) = (12x-18)(12x-16)`

`color(white)(0) = 6(2x-3)(4)(3x-4)`

`color(white)(0) = 24(2x-3)(3x-4)`

Hence roots:

`x = 3/2" "` and `" "x = 4/3`