Question

How do you solve `3x ^ { 2} = 2( x - 1)`?

Answer 1

`x1 = ((2+sqrt(-20))/6) and x2 =((2-sqrt(-20))/6))`

Explanation:

Given:
`3x^2 = 2(x-1)`
`3x^2 = 2x - 2`

Rearranging this gives us a Quadratic Equation, that is;
`3x^2 -2x + 2 = 0`
`x = (-b+-sqrt(b^4-4ac))/(2a)`
`x1 = (-(-2)+sqrt((-2)^2-4(3)(2)))/(2*3)`
`x2 = (-(-2) -sqrt((-2)^2-4(3)(2)))/(2*3)`
Therefore, `x1 = color(red)((2+sqrt(-20))/6) and x2 = color(red)((2-sqrt(-20))/6))`