How do you solve 3x ^ { 2} = 2( x - 1)3x2=2(x1)?

1 Answer
Apr 22, 2018

x1 = ((2+sqrt(-20))/6) and x2 =((2-sqrt(-20))/6)) x1=(2+206)andx2=(2206))

Explanation:

Given:
3x^2 = 2(x-1)3x2=2(x1)
3x^2 = 2x - 23x2=2x2

Rearranging this gives us a Quadratic Equation, that is;
3x^2 -2x + 2 = 03x22x+2=0
x = (-b+-sqrt(b^4-4ac))/(2a)x=b±b44ac2a
x1 = (-(-2)+sqrt((-2)^2-4(3)(2)))/(2*3)x1=(2)+(2)24(3)(2)23
x2 = (-(-2) -sqrt((-2)^2-4(3)(2)))/(2*3)x2=(2)(2)24(3)(2)23
Therefore, x1 = color(red)((2+sqrt(-20))/6) and x2 = color(red)((2-sqrt(-20))/6)) x1=2+206andx2=2206)