Given the equation: (x3−3x)−2=0 transform (−3x+(x3+1))−3=0 or (−3x+(x3−(−1)3))−3=0 −3(x+1)+(x3−(−1)3)=0 (x+1)((x2−x)+(−1)2)−3(x+1)=0 Take common factor 1 + x from the equation we get: (x+1)(((x2−x)+(−1)2)−3)=0 or (x+1)(x2−x−2)=0 then: x1=−1 and also we get the equation x2−x−2=0 This equation is of the form
a*x^2 + 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 = b^2 - 4*a*c - it is the discriminant. Because a=1 b=−1 c=−2 , then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (1) * (-2) = 9
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or x2=2 x3=−1 The final answer for x^3 - 3*x - 2 = 0: x1=−1 x2=2 x3=−1
Vieta's Theorem
it is reduced cubic equation px2+qx+v+x3=0 where p=ab p=0 q=ac q=−3 v=ad v=−2 Vieta Formulas x1+x2+x3=−p x1x2+x1x3+x2x3=q x1x2x3=v x1+x2+x3=0 x1x2+x1x3+x2x3=−3 x1x2x3=−2