Question

Apply Cardano's method for finding the roots of 2x³+3x²-8x-12=0

Answer 1

$$2x^3+3x^2-8x-12=0$$ $$x^3+(3/2)x^2-4x-6=0$$ $$(ax^3+bx^2+cx+d=0)$$

Substitute

$$x=y-(b/3)=y-(1/2)$$

$$(y-(1/2))^3+3/2(y-(1/2))^2-4(y-1/2)-6=0$$

$$y^3-(19/4)y-(15/4)=0 (y^3+py+q=0)$$

Formula Cardano:

$$Q=(p/3)^3+(q/2)^2, A=((-q/2)+Q^(1/2))^(1/3), B=((-q/2)-Q^(1/2))^(1/3)$$

$$y1=A+B, y2=(-(A+B)/2)+3^(1/2)(A-B/2)i, y3=(-(A+B)/2)-3^(1/2)(A-B/2)i$$

$$Q=-5959/64$$

Since the value is less than 0, 3 complex numbers will be solutions of the equation

$$A=(15/8+(5959^(1/2)/8)i)^1/2$$ $$B=(15/8-(5959^(1/2)/8)i)^1/2$$ $$y1=(15/8-(5959^(1/2)/8)i)^1/2+(15/8+(5959^(1/2)/8)i)^1/2$$

$$y2=(-(15/8-(5959^(1/2)/8)i)^1/2+(15/8+(5959^(1/2)/8)i)^1/2)/2+3^(1/2)i*((15/8-(5959^(1/2)/8)i)^1/2+(15/8+(5959^(1/2)/8)i)^1/2)/2$$

y3=(-(15/8-(5959^(1/2)/8)i)^1/2+(15/8+(5959^(1/2)/8)i)^1/2)/2-3^(1/2)i*((15/8-(5959^(1/2)/8)i)^1/2+(15/8+(5959^(1/2)/8)i)^1/2)/2

Answer x1,2,3=y1,2,3-1/2