How can I solve this: a(3x {}^(2) + 2x + 1) = 4x - 6x {}^(2) - 4 to get : x1 + x2 = - 6

Question

How can I solve this:


`a(3x {}^(2) + 2x + 1) = 4x - 6x {}^(2) - 4`
to get :

`x1 + x2 = - 6`

Answer 1

Hey !

The Sum of Roots is -6

Manipulate the Equation to Get :


`(3a + 6) {x}^(2) + (2a - 4)x + (a + 4) = 0`

Now, by Viete,


`- 6 = ((4 - 2a))/((3a + 6))`


`18a + 36 = 2a - 4`


`16a = - 40`


`= > a = ( - 5)/(2) = - 2.5`