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

1 Answer
Xodiac M.
Apr 22, 2018

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

Explanation:

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

Rearranging this gives us a Quadratic Equation, that is;
3x^2 -2x + 2 = 03x2−2x+2=0
x = (-b+-sqrt(b^4-4ac))/(2a)x=−b±√b4−4ac2a
x1 = (-(-2)+sqrt((-2)^2-4(3)(2)))/(2*3)x1=−(−2)+√(−2)2−4(3)(2)2⋅3
x2 = (-(-2) -sqrt((-2)^2-4(3)(2)))/(2*3)x2=−(−2)−√(−2)2−4(3)(2)2⋅3
Therefore, x1 = color(red)((2+sqrt(-20))/6) and x2 = color(red)((2-sqrt(-20))/6)) x1=2+√−206andx2=2−√−206)

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