Given the equation: x3−2x2+2x−4=0 transform x3−2x2+2x−4=0 or x3+2x−12=0 x3−2x2+2x−4=0 (−2x+4)(x+2)+(x−2)(x2+2x+4)+2x−4=0 Take common factor x−2 from the equation we get: (x−2)(x2+2)=0 or (x−2)(x2+2)=0 then: x1=2 and also we get the equation x2+2=0 This equation is of the form a∗x2+b∗x+c=0 A quadratic equation can be solved using the discriminant The roots of the quadratic equation: x2=2aD−b x3=2a−D−b where D=b2−4ac is the discriminant. Because a=1 b=0 c=2 , then D=b2−4∗a∗c= (−1)1⋅4⋅2+02=−8 Because D<0, then the equation has no real roots, but complex roots is exists. x2=2a(−b+D) x3=2a(−b−D) or x2=2i Simplify x3=−2i Simplify The final answer for (x^3 - 2*x^2 + 2*x - 1*4) + 0 = 0: x1=2 x2=2i x3=−2i
Vieta's Theorem
it is reduced cubic equation px2+x3+qx+v=0 where p=ab p=−2 q=ac q=2 v=ad v=−4 Vieta Formulas x1+x2+x3=−p x1x2+x1x3+x2x3=q x1x2x3=v x1+x2+x3=2 x1x2+x1x3+x2x3=2 x1x2x3=−4