Question

How do you solve `3x^2+2=4x`?

Answer 1

`=> x= 2/3+sqrt3/3i" "` and `" "x=2/3-sqrt3/3i`

`x!inRR" but "x in CC`

Explanation:

If able to do so the quickest way is to factorise.

Subtract `4x` from both sides giving:

`3x^2-4x+2=0`

Try 1
`(3x+2)(x+1) = 3x^2+3x+2x+2 color(red)(larr" Fail")`

Try2
`(3x+1)(x+2) =3x^2+6x+1x+2 color(red)(larr" Fail")`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Looks as though we need to use another approach

`color(blue)("Completing the square")`

For detailed method of approach look at
https://socratic.org/s/awA8fpNk

Given:`" "3x^2-4x+2=0`

Write as: `3(x^2-4/3x)+2+k=0`

Giving:`" "3(x-4/6)^2+2-4/3=0`

`=>3(x-2/3)^2+2/3=0`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`(x-2/3)^2=-2/3xx1/3 -= -1/3`

`x-2/3=sqrt(-1/3)`

Square root of a negative number means that the curve does not cross the x-axis so there is no solution in the set of numbers called 'Real'. Written as `x!inRR`

However there is at least 1 solution in the set of number called 'Complex'. Written as `x in CC`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So we now have:

`x=2/3+-1/sqrt(3)sqrt(-1)`

`=> x= 2/3+sqrt3/3i" "` and `" "x=2/3-sqrt3/3i`

Tony B